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0:01
So this is a review of our entire differential equation course in the undergraduate level so this is the part one of that so we can start with the basics so uh what we mean by a differential equation an equation an equation an equation with at least one derivative with at least
0:28
at least one derivative at least one derivative ah is called a differential equation is called a differential equation is called a differential ah equation ah for example so these are two differential equations so d y over d x here x is the independent variable plus x equal y so y is the dependent variable so x is the independent variable so that is normally what is in the denominator ah and then y is the dependent variable what is the dependent variable
1:27
so when you have a ordinary differential equation there's only one independent variable if there are more than one independent if there are more than one independent variable we call it as a partial differential equation we can talk about that later but that's not the uh topic of this course good uh so another example uh so for example this one three 3 d squared y over dx squared plus 5 d y over dx minus y equal 3 so this is another differential equation ah because there are there are there's a
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second derivative and the first derivative and then this is an equation so it's a differential equation okay and then the next thing is what we mean by an initial value problem so the next one is initial initial value problem so in short we call this is i v p so when you see ivp uh that means initial uh value problem so what's the initial problem initially problem is a differential equation and initial condition so initially problem means a differential equation
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differential equation plus initial condition so if you are given ah these two information and ask you to solve the problem we call it initially problem so initial value uh it can be more than one value uh depending on the order of the differential equation if it is first order one initial condition so let's try that so if it is first order we need one initial condition
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and if it is second order we need two initial conditions
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so initial means it given at the starting point so for example uh let's say there's a system that can start at t equals zero so if the information given at t equals zero we call it the initial condition so normally like it looks like y zero or x 0 if x is the dependent variable and then there's another type of problem what we call boundary value problem so what happened it can give you values not just at the beginning but some other point
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so just write it just for completeness so what we call boundary value problem boundary value a problem uh in this course actually we did both so it's called b v p boundary value problem so it's a b v p ah that means uh so you can see the conditions not just at zero but at some other point like maybe like t equal b or something so you can say for example a y b equal something like why not so that is what we call a boundary value brown boundary condition boundary
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condition so initial condition looks like y zero equal something like y one yeah that's not really the right notation but yeah something like that so if there is a case so if you get a boundary value then you we call it's a boundary value problem if you have an initial value equals initial value problem like that and then depending on the degree of the depending on the degree of the differential equation you need to give that many conditions all together so if it is first order you
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need only one condition with second order you need i should say at least two currencies but two is like yeah two is the minimum so if the nth order differential equation you need in many conditions okay so that's that's throughout the course are we gonna use uh that fact good now let's uh focus on the topics so we're gonna go one by one and this can reflect the lectures so in the description you're gonna see link to the entire lecture so part by part so that
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if you have some trouble understanding certain topics you can uh refer the lecture okay so let's start with the first topic so we're gonna go with uh what we call the uh the first one what we did was like verifying solutions so let's start with that so the first topic is we're going to do verifying solutions
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uh verifying solutions so what happened here we are not going to find the solution you are given some expression and you're going to check whether that is a solution so that's the idea and we're going to solve the problem later so let's look at a problem like this uh so um an example like that so let's say verify verify that so remember we say we are not say find we say verify so the solution is given okay so it's a different problem we are not going to find it
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we just verify so we need to understand the wording verify that x equal c 1 e to the minus 3 t plus c 2 e to the 2 t is a solution is a solution of x double prime plus x prime minus 6 x equals 0. so first verify that that's a solution so what we normally do is we take the derivatives and plug into the left side and simplify and so that you get zero that means it satisfies the differential equation that means that it's a solution so that's so what you did was we verify
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we did not find it actually we know how to find this one later when you talk about second order constant coefficient differential equation but that's not the point here we just verify that that's the solution okay because sometimes you can guess so that's the point especially when there are very difficult problems okay so find ah and then we can say find c1 and c2 such that such that normally we put s dot t dot but right now it's the first time
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uh here so i'm going to write such that but later it's be s dot t dot okay such that um such that x 0 equals 0 and x prime 0 equal negative 10 it can be just some numbers it doesn't really matter how you do it but let's see how we can do this okay so the answer ah let ah x t it's a i'm going to exaggerate that as a function of t uh c 1 e to the negative 3 t plus c 2 e to the 2 t so we're going to use that and then what we can do are taking derivatives so taking derivatives
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taking derivatives we have x prime t equal so we enter the derivative say e function what happened so let's recall the results from calc 1 if you take a derivative of an exponential function with the constant here what's going to happen the constant come out and nothing happens to the rest of the function so that's the result so k e k t so k has to be a constant okay if not then you have to use the chain rule okay so we can use that fact here so that means negative 3 c 1 e to the negative 3
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t plus the same thing 2 going to come out so e to the 2 t we need the second derivative as well so let's find the second derivative as well so the second derivative is we multiply by another negative 3 that means become 9 c 1 e to the negative 3 t no change plus this become 4 or c 2 e to the 4 t oh see we can make mistakes very easy so be careful this is 2t no change good uh so now what we gonna do is we are going to substitute so we simply substitute that this in the
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difference liquidation by substituting by substituting
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ah into the equation so what we're going to do we're going to start with one side okay uh
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into the equation x double prime plus x prime minus 6 x equals zero i think under we're gonna start with the left side and see what's gonna happen so let's start with the left side so x double prime plus x prime minus six x equals so let's substitute x prime x double prime which is 9 c1 e to the minus 3 t plus 4 c 2 e 3 t that's x double prime and x prime which is negative 3 c 1 e to the negative 3 t plus 2 c 2 e 2 t that's the uh x prime and then minus 6x so that means you have to multiply
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x with the minus sign so that means a minus so let's multiply by minus 6 so it's minus 6 c 1 e to the negative 3 t and then minus 6 rc 2 e to the 2t now what we can do we're going to see whether you're going to get 0 so let's see so let's go with e to the negative 3 t terms if you look at e to the negative 3 t terms we have nine here we have nine and we have negative three and we have negative six and you can see that when you add them they cancel out so they all those three
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cancel out now so that's gone and now ah actually there's a small mistake so this has to be uh 2 good now let's look at the two t terms so if you look at the two t terms and you can see that we have 4 we have 2 and we have negative 6 and you can see that they all get cancelled out so that's also cancelled but i'm going to cancel it in a in the other direction so that you know what you did so don't just cancel in the same direction one you cancel ah
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this positive other one cancel negative negative you know with the slab slope so then you know what you did uh so that means they all cancel out so you simply get zero so it simply cancel out and you get 0 that means um your expression is actually a solution of the differential equation so we can write that so therefore so you can say that therefore that's how you're going to complete the argument so uh therefore ex like we're doing mathematics means we are making a logical
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argument so it's an argument so like you saw the reasons and said conclusion the reason and the conclusions is like that so it's a nice argument okay um so let's try to do it like that it's like much better if you want to do that like okay so therefore x equal c 1 e to the minus 3 t uh plus c 2 e 2 t is a solution is a solution of the differential equation of x double prime plus x prime minus 6 x equal 0 so we prove that
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yes is a solution now what we can do now we can use the initial conditions to find c1 c2 so using initial conditions using initial conditions using initial conditions so we have x 0 which is uh uh what what you do is you plug in to the independent variable you plug in 0 so that means see uh x0 so wherever you see t plug in 0 so c1 e to the negative 3 0 plus c 2 e to the 2 times 0 so that means we know that e to the 0 is 1 so we know that fact we know that e to the 0
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is 1. so we can use that fact and then that means you simply get c1 plus c2 but we know that from the condition it is simply 0 it's given that x0 equal 0. so we um we use that condition here okay so this is given this equals 0 is given now what else we need is we're going to use derivative condition so we know that x prime 0 which is negative 3 c 1 e to the negative 3 0 plus 2 times c 2 e to the 2 0 now again same argument so you will get negative 3 c 1
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plus 2 c 2 but we know that uh is given that this is equals to negative 10 again this is given okay so it's not our choice it's given okay now the goal is to solve these two equations okay so uh so we have these two equations c one plus c two equals zero negative three c one plus 2 c 2 equal negative 10 now i'm going to call these two equations are 1 and 2 because it's going to make it easy so maybe you learn how to solve these equations when you're in high school or
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middle school so there are two methods to solve one is you're going to solve solve for one and substitute that in the other one i do not recommend that recommend that method because that's that's long pretty long for these kind of things we are not going to do that what we can do uh we gonna try to get matching coefficients but that's much faster and there's a video just to um learn like how to solve these kind of equations and you can learn more about that uh if you wanna watch that and
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even like how to go for two three like that so we can use that technique here the matching matching coefficients technique so what's the matching coefficients what we're gonna do and you can see that especially when you see positive or negative we try to go for that it's much quicker so you can see that in the bottom equations you're going to see negative 3 here so what you're going to do in the first one also you try to get a 3 but with opposite signs so for example you target 3 c 1.
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so if you get 3 c 1 in the top one now what we can do we're going to simply add them once you add them you can see that c 1 term going to cancel out so you simply get c 2 so that is the fastest way to solve this one so that's what we're going to do what we call matching coefficient technique okay so what we're going to do by uh 3 times the first one plus the second one if you do that because if you multiply by 3 you get 3c 1 plus 3 c 2 when you add c 1 get cancel you simply get 3 c 2 plus 2 c
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2 equal negative 10 and then this says that 5 c 2 equal negative 10 now what we can do we can divide both sides by 5 so you get negative 2. so you can see it's very easily okay you get the answer and then what we can do you can pick one of the equations by ah but the first one is easy so by 1 you can see that c 1 is simply negative c 2 so just plug in the c 2 value here so it is negative negative 2 that means 2. you don't want to show all these steps but since this
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is the first example of this type i'm showing all these steps but later we can ignore uh some of them some obvious steps so we're done now we're going to complete that so someone make an order so you're gonna preserve your uh results in the best possible way okay that's how you're trying to do that's why we're writing all those words okay to make a good argument so therefore um therefore we can say the solution the solution to the initial value problem to the initial value
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the initial value problem is is we can write that x t which is equals to 2 e negative 3 t minus 2 e to t so that is the uh final answer for that problem so what you can see that you know if you to do this it doesn't take much time so it's like maximum it would take like 10 minutes to do all of this but i'm doing it slowly just to explain everything but you know when you want to do that in exam or something you don't want to like we
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don't explain like i mean we we explain but not like we are not going to take too much time just right fast okay so it will it won't take too much time you can see you can practice and see how long it takes for a problem like that okay good so that's the first problem uh and if you need more you can look at the first video about that to find more information and what else we did okay good so let's go to the next topic so the next topic is what we call the uh four fundamental equations
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so when you talk about four fundamental models so let's talk about that for fundamental model equations for fundamental for fundamental model equations so we talk about four models and why we did that because we know that these models like appear pretty much everywhere in sciences so it's good to know those four separately ah okay so uh the first one uh what we did was the exponential growth model so exponential growth so that's pretty much every guys like
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you know bacteria even like these days kovid all of them follow the exponential growth model um and then uh so the differential equation is d y over dt we can use the t as the uh independent variable as a time it's my more it's like kind of makes sense so k is a constant we assume k is positive so and then we know that the solution of this one is we can write as y t equal a constant times e to the k t so that's the solution and you can
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remember you can use this you know you know that even we can we can solve this very easily uh just uh this is like a separable differential equations or we can even use a simple coin technique to solve this one okay it's not a difficult uh difficult problem okay so now uh if you look at the uh the solution for the graph of this one so uh what is the graph of this exponential growth that means you can start at a value and then it's going to grow ah so this is what we call the exponential
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growth model that's the starting value a okay and then it's going to grow so we assume that k is positive it's a growth okay that's the first case and then we talk about the uh what you call exponential decay ah so exponential decay
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exponential decay model source exponential decay is very similar d y over d t and then we assume that k is positive but with the negative sign in front like that so it's pretty much similar what's going to happen there's a negative sign in front so the e negative k t so that's what happened and uh so if you look at the graph uh very similar but what's going to happen now is it's going to decay so you can start with the value and go down so uh so like that you can see start with the value a and
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then go down so you can see this exponential decay and we assume that k is positive again because that we put a negative sign in the equation so that's particularly exponential decay especially like if you look at radioactive material it's going to follow this kind of pattern uh there are a lot of other things like even bacteria if you kind of uh put like a disinfection there and those bacteria are gonna um kill and then uh if you look now if you
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try to count so it's gonna follow something like that so this this this models everywhere like you know if you get injection what's gonna happen so so those are the kind of models for that so it's very common situation it's called exponential decay and then we talk about the next most common one is what we call the simple harmonic motion so simple harmonic simple harmonic motion sometimes we call shm okay simple harmonic motion
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uh so what is the differential equation there's a slight difference now this can be a second order difference frequency we have like d squared and then we that we're going to use the time variable is a time equal um so we can write negative there's a negative sign and a k squared y so remember earlier you had a t now this is y so that's the major difference here that's a dependent variable comes here and then we know the solution so the solution is with sine cosine
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so some number and then we have cosine k t so k is the square root of that term so you see you take the square root of that and then another constant uh sign so one cosine it doesn't matter which one we write first uh so like that so that is the solution so it's a sine cosine we know the sine cosine are periodic functions so that means if you look at the graph it has a graph like this so it's going to be periodic and it's going to fluctuate
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start with some value depending on the initial condition is gonna go in a uh sine curve or cosine curve because it's a mixture of that so yeah so that's what happened uh so this what we call a simple harmonic motion this is everywhere like all the atoms um in the body uh follow these kind of motions and if you have a mass and a spring uh it's gonna follow these kind of patterns so this is like kind of very common uh so it's called the simple harmonic motion and then uh the fourth one what we
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normally call the hybrid model so we have a hybrid model uh what's a hybrid model is we have the experience of growth and the exponential decay you can combine them into one that's a hybrid model and you can see the interesting relation it is simply the same equation we just write above but with the plus sign so that's very interesting what happened there's no t here it's the plus sign the above one but with the plus sign we stick with the k squared so what's going to happen we can write the
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solution of this one in two different ways uh the one way is using the hyperbolic cosine hydraulic sine functions of cos kt plus sin uh kt able h to mean that's a hyperbolic functions and then if you are not familiar with this you can look at the uh the video for that we explain that and we also talk about some other properties for example derivative one of the major thing is that when you take the derivative uh cost becomes shin and shim become cos there are no negative signs that's one
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of the major difference between the regular sign in question also uh you can write this solution in different way uh i i'm gonna still use a b but those are not simply be some other constants but i'm still gonna say a b so uh e to the uh we can write one of them let's say k t as a k t and then plus b e to the negative kt that's why we call the hybrid model uh this is the exponential um growth model and that's the exponential decay model so you add the two and you get the
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hybrid model so that's the hybrid model for uh when you have that different sequence that's hybrid so those are the four fundamental model equations and if you look at the graph uh so you're going to get like graphs like this which i'm going to put both together uh so you can get a graph like that so because it's a part sine part part of e to the kt part in the minus kd so whoever the win gonna get the uh formula so it's gonna be like that uh oh like you know it can be like that
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okay because we know that uh if you hang a wire heavy wire it's gonna go in a uh equation like that uh so it's like in many situations and so uh yeah good now let's try to see how to solve a problem using uh these models so the goal is if you see one of them you can directly write the solution okay because these are familiar one remember this this is very useful for your other courses so you can directly use the results so let's look at a problem so let's solve this problem uh so we have
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let's say solve the initial value problem d squared x over dt squared equal minus 4x along with x0 equal let's say 12 x prime 0 equals 0. uh why we need two conditions because this is a second order differential equation so you need two conditions now what we can do we're going to match this equation with one of those four models and you can see that this is the simple harmonic motion situation and you can see that uh in this problem k squared equal 4 this implies k equals 2 because we
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consider only the positive value now we can write the general solution so the general solution is so the general solution is uh you can say x t equals you can directly write it it is a cos 2 t the 2 is t is the independent variable you have to be careful so the independent variable goes there okay and then also this x and this has to be x otherwise it is not this type of problem good um and then so we have uh plus plus b sine 2t so we have that so that's the it's
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directly come from the four model equations now what we can do are we going to use the initial conditions but we can see that when you want to write in use the initial conditions that's a prime so that means let's take the derivative first so taking derivatives taking derivative so we can write x prime t equal so what's the derivative of this sum it's a cosine so derivative means the negative sign but there's a 2 here that
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means by the chain rule that 2 comes out so that's going to be negative 2 a sine 2 t plus same thing sine when you take the derivative it's going to be cosine so with me 2 b cosine 2t by the chain rule now we can use the initial conditions so um using initial conditions using initial condition uh we have uh x0 x0 that means plug in 0 for the t so you're going to get a cosine 0 plus b sine 0 which is equals to 12 this is given now we know that cosine 0 equals to 1
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sine 0 equals 0. so because of that this simply implies that a times 1 that means a plus 0 equal 12 that means a equal 12. so we find one value now using the other condition we know that x prime 0 which is equals to the second one so negative 2 a you can actually put the value here so it is we already know the value so i'll put the value here just to make the calculation easier so 12 ah sine 0 plus 2 b cosine 0 which is equal to 0 so again we know that the sign 0 equals 0 uh cosine
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0 equals to 1 so that means this says that so nothing comes from the first term this says that 2b equals 0 so this says that b equals zero that means the b term gonna go away that means that the solution is therefore the solution to the initial value problem is simply x t equal 12 cosine 2 t so that's the solution of the problem good um so let's go to the uh the next topic so what's the next topic so as we gonna do uh we're gonna talk about uh the classification of differential
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equations so um so number three are the classification of different equations
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classification of differential equations so we talk about various types so we first start with what we mean by ordinary differential equation
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ordinary ah differential equation differential equation so we normally call this ode ah in this one what we have one independent variable one independent one independent a variable
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so we have one independent variable so ah an example would be something like this dx over dt r plus 3 x equal t plus 1 you can see the independent variable is t and that's there are no more independent variables so these are ordinary differential equations o d e and then the next type is what we call the a partial differential equation so partial differential equations
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partial differential equations
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partial difference equation or we call this is a p d e there's a certain form for that p d e partial differential equations so what we have more than one more than one ah independent variable
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more than one independent variable so there are more than one independent variable sometimes the space and time those are the two in two common independent variables so uh so such a differential equation have a kind of form like this so we can write let's say u is a function of x and y so then we can write it like this so if you use a function x and y if u uh is a function of x and y so that u is a function of x and y sorry right so then you can write a differential equation uh like this
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uh you can see we can use the dava for that so double u over del by x plus three times w over double t equal we can say x uh plus t so this is a partial differential equation and you can do that there are two independent variables space and time so x and t are independent variables but that is a different course and we are not going to focus that in this course this is only for a partial differential this is only for ordinary differential equations except in few cases because we need uh
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the idea of partial differential equation in in certain situations and other than that we are not going to focus on that okay that's about that so there are two uh independent variables so these are the uh independent variables independent okay so there are two in the first one you only have one okay um and then we talk about what we mean by a order of a differential equation so that's the next topic we talk about the order of a differential equation
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so order so order means the highest derivative appear highest derivative highest derivative highest derivative is what we call the order so for example if you look at this differential equation d square d squared y over d x squared plus x equal three so you can see the highest derivative appear is two so we call this is a second order difference liquids we don't see the two order so we say this is a second order uh ode second order ode good uh that's the highest they really
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appear the question why do you do that because you know that uh depending on the order we had to use different techniques that's just a major point here so the next important topic is what we mean by linear and non-linear equations so the next one is uh before a linear and non-linear so linear versus nonlinear
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ah so that means linear means no powers other than one of the dependent variable so you're going to see no powers no powers other than one
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other than one of the dependent variable
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of the dependent variable so we don't worry about the independent variable so there can be only powers of one so that means there are no products okay in other words no products no products of uh the dependent variable
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dependent variable or its derivatives so let's talk about two examples uh if you say x is the dependent variable so we can write dx over dt plus t cube x equal t so you can see that there are no powers of x other than one so this is a linear equation t3 t is the independent variable that's okay so this is linear and if you look at this one x times d x over d t plus the same thing here t3x equal t this is non-linear
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so those two equations make a big difference when you try to solve linear equations are easy to solve in equations are easily solved so that's the fact linear equations
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are easy to solve
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good ah so these are linear non-linear good and then talk about what we mean by the constant coefficient so number five constant coefficient constant coefficient constant coefficient uh equations so that means all the constants are coefficient all the constants all the um coefficients or i'm sorry all the coefficients are constants okay all the coefficients are constant so for example five x double prime minus three x prime plus x equals zero so this is a constant
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coefficient differential equation you can see that all the coefficients are just numbers uh what we normally do is we consider a characteristic equation to solve okay so uh so use characteristic equation uh use a characteristic equation to solve which remember which which is in lambda so we write a lambda equation great um and there's a characteristic equation for uh matrices as well but here uh we call the characteristic equation for that because that can characterize the
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difference equation okay that's a constant coefficient and then let's talk about homogeneous versus non-homogeneous so the next one so we have homogeneous
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uh versus non-homogeneous
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so be careful are the homogeneous and non-homogeneous has a different two different meaning in differential equations uh and then we can talk about the homogeneous and non-homogeneous again later so here um uh what we mean by the homogeneous is uh whether uh the difference equation equals zero or not so like uh it will be easier from an example so for example if you have five x double prime minus three x prime plus x equals zero you see that ah so you can see that
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ah so you're gonna set the digital equation to 0 that means that side is not a function of the independent variable so all the dependent variable you can put it one side and then you can see that so we can write this one as a this is what you call a homogeneous equation so homogeneous so no functions of the independent variable okay so it's a homogeneous equation now you can say 5 x double prime minus 3 x prime plus x equal t this is non-homogeneous
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because it's a function of ah the independent variable okay so if you have a function because you can always pull all the uh dependent variable terms to one side and then if you see a function of the independent variable alone that's going to be non-homogeneous okay you see a function of the independent variable alone a zero we just say other than zero okay because here is also a function but other than zero so this is what we call non-homogeneous because of that t term so this is
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non-homogeneous
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good ah homogeneous and non-new so there are there's a function of the independent variable alone okay so that means uh let's write that so there is there is a function function of the independent variable only dependent variable only okay so if you see a function of independent variable only it is what we call non-homogeneous good ah and then we have what we call autonomous autonomous
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autonomous differential equations
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so autonomous means if the differential equation does not depend on the independent variable we call it the autonomous differential equation the differential equation differential equation does not does not depend on the
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does not depend on the independent variable independently
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ok if the difference does not depend on the independent variable we call it is a autonomous differential equations let's try this one so you can say dx over dt plus x squared equal 3 so you can see that this is this does not depend on the t variable so this is autonomous one okay so this is autonomous so if you had like a function of t somewhere so it is it is not autonomous it's autonomous okay and then the last topic of this is what we call system of worries
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system of audis so a system of od so not just one equation but there are many equations so like for example x prime equal 2 y plus e to the t y prime equal let's say 3 x plus y sine t so this is what we call a system of orient ah so we discuss all of these types good uh so that's about that ah so i don't discuss any further examples of this one but if you need more examples you can look at the corresponding lecture for that but this is the kind of
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basic ideas we talk about let's go to the next topic uh so the next one is what we call the basic integration so number four basic integration
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basic integration so this is the very simple thing that we did we did even this one in calc one ah so there are two types so let's go with the uh type one
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so what's the type one you have a differential equation which is just on the on the right side it is just a function of the independent variable so this is a function of the independent variable so this is a function of the independent variable only
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independent variable so if this is a function of independent variable things are very easy just integrate the two sides so that means y is simply the derivative of a y simply the integral of the derivative okay we just gonna integrate with respect to x so that they cancel each other so that means simply become the integral of f x dx it's very simple just integrate uh that's gonna happen if the derivative is a function of the independent variable only so this is the
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easiest case you can simply integrate there gonna be careful uh because sometimes um it's not the case okay so let's talk about type two so what's the type two is
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so type 2 ah we have d y over dx but it is a function of the dependent variable now okay so if you get the function of the dependent variable so let's write so what happen now is is a function of function of the dependent variable
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so what we can do in this case we can flip the equation over so we can write this on as dx over d y 1 over g y now you can see that this came back to the previous case because you can simply consider y as the dependent variable in this case so we can see consider
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consider y as the independent variable so that's the whole point and then just indicate the two sides again with respect to y now so that means you're going to get x which is the integral of dx over d y with respect to y not x integral of g y d y so that's what's going to happen in that case we're going to get with respect to y okay those are the the the first type what we call the basic integration so you can just use a table of integrals for that
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uh the common ones and you can do a lot of problems just in this these two equations so the first case is the uh derivative is a function of the independent variable second one is the derivative of the function of the dependent variable in the second case we're going to flip the equation over and integrate so let's do two examples from those two types so we have this first example so let's say uh solve
51:33
the initial value problem is say dy over dx equal 3 cosine x 3 cosine let's say 2x minus 8 e to the negative 4 x plus 5 along with y 0 equal negative 3. something like that and what do you notice you notice that this independent variable this is the independent variable and you can see that it's only a function of the independent variable so that means this is the case one that means you can simply integrate the two sides ok so you can see the function of
52:22
so this is a function of x okay x is the independent variable so that means we can have we can integrate so by integrating just integrate
52:39
okay by integrating y equal just the integral so it is 3 cosine 2x minus 8 a to the negative 4x plus 5 dx so this is just very basic integrals and if you uh forget so these are the kind of formula that we're going to use here so if we have e to the k x dx the answer is simply 1 over k e to the k x plus c this only works when k is a constant and the second one if we have a cosine k x r d x and we know that it is simply 1 over k sine kx so just divide by k and the other one if you have a sine
53:36
kx k has to be a constant so this is simply negative 1 over k cosine kx plus c so those are the kind of three common ones and then also we have the power rule that everyone know and so this is what we're going to use now so that means in this case k equal 2 so simply divide by 2 this is a cosine that means it's a sine one so divided by 2 so that means it's going to be 3 halves sine 2 x minus ah you can see 8 you have to divide by negative 4 e to the negative kx plus
54:19
5 is a constant size of ix plus c so that's what you get and you can simplify this a little bit so you're going to get 3 halves sine 2 x simplify you get 2 e to the negative ah this k actually 2. okay actually 4. so you get 4 ah so e to the 4x plus 5x plus c now what we can do is we can use initial current condition so using the initial condition y 0 equal negative 3 you can see that negative 3 equal y 0 that means wherever you see the independent variable plug in 0
55:07
so you're going to get 3 halves sine 0 which is 0 plus 2 e to the 0 which is 1 plus 0 plus c so you can see that this is 0 and this is 1 so this tells you this tells you that negative 3 equal 2 plus c so in other words c equal negative 5 so you can plug in that so that means the solution is simply y equal three halves sine two x plus two e to the four x plus five x minus five so that's the solution so you can see that it's not difficult uh we just use the uh basic integration
56:09
technique because the right side is just a function of the independent variable now let's look at the other type so what's the other one this example so we have let's say solve the ivp the initial value problem e y over d x equal y plus 3 along with y 1 equal negative 2 something like that so if you see a equation like that now what we can do you can see uh this is a function of the dependent variable so this is a function of the dependent variable
57:00
so when you see function dependent variable what you do you're going to flip this over also you can do this problem using the separation of variable actually that's a separation of variable but you can see as a technique you can remember like that uh so we're gonna flip this over so we could write this on dx over d y equal one over y plus three now what we can do integrate with respect to y so integrate
57:32
integral with respect to y now so if you do that you get x here and it's simply the integral of 1 over y plus 3 d y so you can see there's a 1 over y plus 3 3 is just a constant so this is a log function so this simply becomes natural log five plus three we are not going to use the substitution here uh that's like long time ago okay so we have the solution now so what we're gonna do is uh we can find the uh constant and then uh so
58:09
i mean there are two ways to do it first we can solve for y and then plug in the constant or we can plug in find the constant and then we can solve uh so so let's do that so we know the constant uh so what it says that so the using the initial condition
58:32
okay so there are two ways to do it either you can use the condition now and then go for the uh solve for y or solve for y and then find the constant so uh work both be be careful uh what do you get okay because sometimes when you try to uh write the log form with the exponential form remember you get plus or minus so be careful with that whether that is the case so let's solve for y first so what we're going to do now is you can write this one as ln y plus 3 equal x minus c
59:12
so this says that y plus 3 we have solved many problems like that so you can see that this is absolute sign so that means uh you're going to get the uh plus or minus here uh so we have uh it's a plus or minus ah you can see uh e to the x minus c so uh so now what we can do we can write this one as a to the uh like plus or minus e to the minus c and then a to the x what i'm going to do i'm going to call this whole thing just a c so if you do that or some other letter
1:00:00
if you want just what we normally call the by abusive notation i'm going to call this whole thing just c it's not the same c but another one okay like that so we can add it like that so in other words y equal c e x minus 3 now we're going to use the initial condition so using the initial condition we know that uh for the initial condition it's a negative 2 when you plug in x equal 1 so let's do that so that means negative 2 y equal negative 2 equals c e negative 1
1:00:33
negative 3 and then we can what we need to do is we need to solve for c so let's do that how to do that first are we going to move the number to the other side that means we can add 3 to the both sides so this says that 1 equal c e negative 1 in other words if you multiply both sides by e this is the c equal e so if you plug in c equal e uh what's going to happen y equal e times e to the r x minus three okay ah so and then you can combine them uh because e means e to the one
1:01:17
so if you combine that uh so you're gonna get y equal e x plus 1 minus 3. so that is the solution yeah i see that there's a small typo here um because we plug in uh negative one it's not negative it's positive so that there's no negative sign so that means uh this gonna be e to the negative one so that means this is negative one uh so that means the final answer this is minus 1 good so that is the solution of the um difference equation and you can see that that's not that difficult but there's a
1:02:01
process that we need to remember good uh so let's move on so let's talk about the next major type uh so what is the what we call the separable differential equations so number five a separable different difference equation separable differential equations so what's the separable difference equations let's look at differential equation we have d y over dx which is a function of x and y but sometimes what happen you will be able to separate them into you will be able to write
1:02:49
this f x y function as a product of function of x and y so if you can write this one as a product of f x and y we call it is the separable equation so this is what we call a separable different equation if we can do that the equation is a separable differential equation separable uh differential equation because you can separate them okay and after we do that what we can do um we can divide both sides by g y so if you divide you can write this one as d y over g y equal on this side we have f x
1:03:27
we're going to multiply we're going to bring dx to the other side so we can separate y's and x terms and then after that what we can do we're going to separate and we integrate them separately so that is what we call the separable equation uh formula okay so we're going to uh integrate them uh separately okay very simple argument uh we're gonna first write as a product and then we're gonna move the gy to the other side by dividing and then integrate the two
1:03:53
sides uh separately so let's look at some examples so example so let's say solve d y over dx doesn't look like x d y over d x equal let's say um y minus 4 x y and then y 0 equal let's say e something so when you get a problem like that um you can check whether you can write this as a product of x and y separately so that's the first step so the answer and you can see that yes you can write the right hand side if you pull y out you can see 1 minus 4 x you can see it's a product of x and y
1:04:48
separately so what we're going to do we can divide this side by the y function leave the x function on this side and then integrate with respect to x on that side integrate this with y on this side so that's the register now what you're going to be integrate the two sides separately first one with x right left side with y right side with x so you're going to get ln uh absolute y here this is just the power rule so you're going to get x
1:05:23
minus if you integrate x you get x squared over 2 so that means you can 2 x squared plus c we are not going to add c 1 c 2 we only add 1 c to one side because you can always simplify now we can write this one using the exponential form so so let's do that ah so if you write them using the exponential form i can do it here then you can see all the work so if you write the exponential uh form so what's gonna happen is y equal the same thing we know that's absolute value
1:06:00
so you get plus or minus and then you can just put one constant so it's the same thing here so it is uh so what happen is a plus or minus and then e to the x minus 2 x squared plus c and then what we can do the plus or minus and then e to the c we're going to combine so if you uh you combine that let's write that extra step so if you combine that plus or minus e and this c uh we're gonna combine so that means you're gonna i'm gonna call it a so if you call that a and then you're
1:06:41
gonna get e x minus two x squared we have done this a lot okay so that's plus or minus e to the c you could just call a and we still have e to the x minus 2 x square good so that's the solution now what we can do we can use initial condition so the using the initial condition that y 0 equal e so we're going to get e equal a e to the 0 e to the 0 means 1 so that means a so a equal e so that means y equal e to the 1 ah we have e x minus 2 x squared this means that's e to the
1:07:29
x minus 2 x squared plus 1 so that is the complete solution it's not difficult okay first we find the general solution that's the general solution
1:07:46
and then we use the initial condition to find the constant okay um let's just do one more example it is very very important uh topic so how about this example so the next example this is solve
1:08:09
x dx over dt equal e 3 e to the t minus x something like that so let's try to find the general solution of this one so what is the independent variable independent variable is t you can always switch uh flip it and then change that uh so it doesn't matter at the end you're gonna get the solution okay okay so the answer so what we're gonna do we uh use a certain algebra here so we write dx dx over dt uh we can write this one as 3 e to the t times e to the negative x so we can do that
1:08:50
now what we can do we can multi we can see whether we can move x terms to one side t terms to the other side actually we you can see that yes we can do it if you multiply both sides by e to the x uh you're gonna get x e to the x dx and then i'm going to leave 3 here i'll leave e t here and then dt i'm going to move to the other side so that we can do so you can see it is a separable equation now so we can integrate the two sides uh separately uh right side is okay how do the left side
1:09:30
for the left side uh you have to use the integration by parts okay so uh so we're gonna make the table so we're gonna pick u equal x you can do the shortcut you can use the di method uh i'm gonna just use this uh regular method okay just to make sure that okay so dx dv uh so if you are trying to use the shortcut method the di method make sure you practice that first okay so dm method and there's a video and that kind of solved many problems there
1:10:00
so you can practice using that okay and then we take the derivative here sorry du equal dx integrate so you get ex and then what is the integration by parts formula it is uh udv is simply equals to uv uh minus vdu so that is the integral by parts formula so we can use that so by the integration by parts uh formula on this side you simply get uv that means x e x minus v d u uh v means e x d u means d x equal here 3 e to the t plus c we had only see one side
1:10:47
uh and then we can finish the integral here so we get x e x minus e x equal 3 e t plus c uh so the question is can you solve this for x x is the dependent variable can you um because of that x term we cannot uh because there's no way you can pull the x term out because when you have sine cosine exponential log um like that there's no way you can pull the x term out when you have a product like that it's a non-linear okay so this makes it non-linear so we cannot so we can leave it like that
1:11:29
so this implicit solution so this is the implicit solution it's implicit solution of x and if your initial conditions uh what you can do is you can plug in them and then solve for c but otherwise you just have to leave it like that because that's no way you can solve this but that's the implicit solution for the x function very good okay so let's talk about the next important type uh that is the first order uh linear differential equations and we're gonna use the
1:12:01
integra adding factors method that's the next major topic okay you always have questions in the exams good uh so let's talk about that now uh so the next topic this is uh six number six so this is first order first order linear first order linear uh differential equations you have to remember like that first order linear so we use the integrating factor method
1:12:42
integrating factor integrating factor method uh so let's talk about the uh let's recall what we did so we write what we call the standard form so it has to be of this form if it is not you have to bring it to this form before we do anything uh and then d y over d x plus p x y equal q x so that's the kind of format for the first order linear this is what we call the standard form standard form
1:13:17
standard form ah and then so if it is not in standard form you have to bring it to this term form first okay so if not
1:13:30
if not uh bring it to stanford that's the first step brain it uh we need to uh bring it into the standard form and if you don't do that your answer is wrong okay so it's very careful with that let it bring it to the standard form good so let's talk about the uh steps uh so what are the steps of such a problem so the first step is so so we assume that it's in standard form okay that's the standard form so we assume this is the case
1:14:14
uh now what we're going to do uh we're going to find what we call the integrating factor okay so find the integrating factor i which is always e to the p x dx p means everything you're going to be with the sign so we're going to uh calculate that after that what we're going to do are we going to multiply we're going to multiply the whole equation by that so we can write i x times the d y d x plus t x i x times y on the other side we have q x times i and now the magic happen
1:14:58
what's the magic you can see that on the left side this is automatically the product rule for the product y and i so we can write the form that's what what you normally call the form so what the form here is automatically this side becomes the uh product of y and i that's automatic what's the product if you apply the product rule that's what you get and on the other side now we have qx and i now what we're going to do to remove the derivative we can integrate both sides so you can integrate
1:15:37
and once you integrate what's going to happen you get y i term on this side this is just the integral uh q x i x so you can see this is a function of x it may or may not be able to integrate if you cannot we leave the with the integral sign dx and we can exaggerate the plus sign now what we can do we need to solve for y so you divide by i so that means y is simply you're going to divide by item so you're going to get the bracket so so we have inside that we have qx
1:16:18
ix dx plus c so those are the steps uh so let's try to don't remember this as a formula that's just because it's much easier to follow these steps and get this quickly that's exactly the whole point of learning these things okay plug in two equations anybody can do that uh so it'd be better to like kind of follow the steps and get that uh good so let's do uh maybe two examples from uh this is very important so i'm gonna do two examples the first example so example number one
1:16:59
so let's say we have a d y over dx minus 3y equal e to the 3 x sine of 4 x y zero equal negative one fourth
1:17:20
so the answer ah so you can see that this is a first order linear difference liquid is in the right form okay so this is first order uh linear
1:17:37
uh equation so that means we're gonna use the you can see that this is not separable equation you can separate x and y uh so we can use indicating factor method so let's go step by step so we're going to find the indicating factor uh integrating factor
1:18:01
so we're going to say i which is e to the p x dx so let's plug in here e to the uh what is the px you can see px is with the sign negative 3 that's the p so it is negative 3 dx negative 3 you can pull out so that means it's going to be e to the negative 3 x we are not going to add the c uh because we are looking for one solution that all the solution now what we can do we can multiply uh so so you're going to multiply the whole equation by that so it is e to the
1:18:42
negative 3x d y over dx minus um you have 3 e to the negative 3x y equal what we get on the other side it is e to the negative 3 x e to the 3 x sine 4 x you can see it simplifies very well okay so that means if you simplify it properly it's gonna work very well so you're gonna get the sine uh 4x okay that's the gift you get if you did that right it simplifies very well okay ah now we know that this is automatically uh this side is so we can write the form so this is automatically
1:19:30
the derivative of the y and the integrating factor so that is automatic thing on the other side we have sine 4x to remove the derivative uh we can integrate so integrate ah with respect to x x is the independent variable so that's going to cancel out so you simply get y e to the negative 3 x the integral of sine 4x dx this is very easy to integrate k equal 4 so you can divide by that this is sine so it's a negative cosine so this can be negative 1 4 cosine 4x plus c don't forget that
1:20:21
and now what we can do we're going to multiply both sides by e to the 3x so if you multiply by e to the 3x both sides so you're going to get negative 1 4 e to the 3x we're going to measure by e to the 3x the opposite sign okay so it's a negative 3 you multiply by positive 3. so so that they're going to cancel on the left side on the other side you multiply by e to the 3x so you're going to get that times cosine of 4x plus c e to the 3x so that is the solution that's a general solution
1:20:56
okay so this is the general solution now what we're going to do since we know the initial condition uh let's plug in that and try to find c otherwise this is the solution see if i need to find general solution we're done okay so let's next step is we're going to use the initial condition to solve it so using the initial condition using the initial condition uh y 0 equal negative 1 4 you're going to get negative 1 4 which is simply equals to y 0 this is negative 1 4 where
1:21:35
you see x puts 0 so e to the 0 cosine 0 plus c e to the 0 which simplifies to negative 1 4 because we know that the cosine is 1 e is 1 this is also 1 so you're going to get negative 1 4 plus c uh you can see negative plus 1 4 negative 1 for the cancel so this is that c equals 0 so that means that term is gone so you simply end up with y equal negative 1 4 e to the 3 x cosine for x as the solution to the initial value problem good uh so let's try to solve uh one more of
1:22:33
that because this is very important so i'm going to do one more so then as the next example let's try this for the example uh solve let's say t d y over d dt minus 2y t cube e to the negative 2t okay good um is this in the standard form so that's the first question you wanna ask okay first question you wanna ask whether this is separable okay that's much easier actually that's not the first first question is whether this is the basic integration problem this is like d
1:23:21
y over dt equal function of the independent variable only or function of the dependent variable only and you can see that no it's none of them so it's not basic and then ask whether this is a separable differential equation can you separate the x term uh can you separate the y terms and t terms and uh you can see that there's a sum uh and also there's a multiplication there's no common term so this is not separable and then we're going to go for the linear first order integrating factor with it
1:23:55
so when you're going to integrate factor method you ask the question can you ecs in the standard form and you can see that this is not in standard form so what you want to do you have to bring it to standard form so how to do that you're going to divide by t so because you don't need that term so that's the only problem so we're going to divide by that first so let's do it so we do the whole equation by t so that's the first step to do okay so we can write not in standard form
1:24:25
not in standard form not in standard form ah so i divide by by that t term once you do that so you're going to get d y
1:24:55
going to get d y over dt minus 2 over t y equal let's begin t squared e to the negative 2 t okay now this is standard form everything looks good now what are you going to do find the indicating factor so let's do that uh so the first thing we need we divide okay so uh find integrating factor integrating factor so which is i equal e to the integral of p t now p t d t so it is e to the integral of p t which is negative 2 over t you have to pick the whole thing
1:25:42
okay uh dt uh this is very easy to integrate we normally do pull the negativity out this is one over t dt one over t means ln that means e to the negative 2 ln t we're going to bring this as a power so that means we can write this one as a to the ln t negative 2 now from algebra this part get cancelled okay so you're gonna get t e to the negative two or in other words one over t squared it doesn't matter which form you write so that is the integrating factor now
1:26:20
what we're going to do we're going to multiply the the equation our standard form equation using by the integrity factor so let's do that ah so when you do that so we're gonna multiply by uh one over t squared the whole equation so you're gonna get one over t squared d y over dt minus is going to be t cube now y equal what happened here it is 1 0 t squared you what you have here you have t cube e to the negative 2 t so you can see this get cancels very well
1:27:10
ah and then you have e to the negative 2 t so then we can write the form so that this side is automatically uh become the uh so we write the form so this side automatically become d over d t y times the integrating factor other side is e to the negative 2 t we can integrate this now
1:27:42
if we integrate you get y
1:27:47
times 1 over t squared this is the integral of e to the negative 2 t dt this is a very simple integral you're going to divide by the constant so that means it is negative one half e to the negative two t plus c now what we're gonna do uh we're gonna multiply both sides by t square okay so multiply uh by t squared to cancel that so we're going to get y equal you simply going to multiply by t squared so you're going to get negative one half t squared e to the negative two
1:28:26
t plus c t squared so if you know the initial conditions uh we can find c okay so if
1:28:40
the initial conditions are given
1:28:47
if the initial conditions are given ah we can find c so that's the idea uh of using uh the integrating factor method so you can see that it's not that difficult but there are certain steps that you need to uh follow so here what we did was we multiply by one or t squared and we write the form integrate and multiply again by t square to isolate y good uh so next we can talk about uh what we call the substitution methods there are three uh so we're going to talk about one by one
1:29:31
so let's talk about the uh substitution method so there are three substitution methods so we're gonna do uh one by one the substitution methods
1:29:49
ah so we can talk about three so let's go with the first one what we call the direct substitution direct
1:30:04
substitution so what's the direct substitution let's it will be clear from an example so what we normally do is let u equal what we call the inside function uh inside a function so we're going to put the inside function as u and then we can write the differential equation using the new variable and rewrite rewrite the differential equation using the new variable using the new variable
1:30:47
so let's see this in a problem so how about this one let's say uh solve d y over dx equal y uh plus a 4x minus 3 to the square maybe something like that so y 0 equal 4. so let's look at this problem and you can see that if you look at this differential equation um if you try to expand this one you're going to get a lot of terms and it would be very difficult to solve because there are y terms y squared terms x y products and a lot of times uh so and you can see
1:31:30
that uh none of the method that we discussed doesn't work for that so what we can do we can use this substitution technique so we can substitute for this inside functions that's the inside function so let's so that's the starting point so what we're going to do uh so the answer is just the basic substitution we're going to pick so let's say let u equal the inside function y plus 4 x minus 3 so that's the starting point we're going to pick that inside function and then just take the derivative
1:32:03
uh so if you take the derivative so uh taking derivative so taking derivative uh we can write this one as uh y prime or divided by dx uh so d u by dx equal d y by dx on the other side plus 4 because 3 is a constant negative so it's going to go away so that's what you get um and then what you can do is you can solve for divided by dx so this is the divided by dx equal ah du by dx minus 4 so we're going to plug in that here so if you plug in that uh what will
1:32:49
happen to the differential equation so d dy by dx you can replace by that so it's going to be d u over dx minus 4 equal are we going to put the inside function as u so it becomes u squared so that means uh so this is that du over dx is simply u squared plus 4 and you can see that this is a separable equation this is separable so we can just use the separation of variable method to continue so we're going to solve for u first after you solve for u then you can use the substitution
1:33:35
this substitution to solve for y okay just uh solve for y okay so that's the idea so let's finish the argument so we have uh so what we have we have du over dx is equals to u squared plus 4 so you can we can move the u terms to one side so we can write this one as d u over uh u squared plus 4 the other side is simply dx now what we're going to do we're going to get the two sides so if you integrate on this side this is simply going to be tan inverse u because
1:34:14
remember there's a formula uh so what the formula says if you integrate uh 1 over x squared plus a squared uh so this is simply 1 over a tan inverse x over a so that's the formulas is plus c and that's what we can use here so this is one of the formula it's a direct formula so we can use that um and also you can find all these formulas in another video i put like you know all the videos in one one formula uh in one video uh so you can see like kind of all the common uh formulas so you can
1:34:57
check the uh character uh uh youtube list for that from that you can find it good and also put a link uh link to uh in the description so you can find that in the description of this video good okay so that's what you get from the left side from the right side it is simply x plus c that's what you get um so this means that if you take tan of both sides um so this says that u equal 10 x plus c uh and then what are u uh u equal now we're gonna substitute back okay so uh since
1:35:42
uh u equal y plus 4x minus 3 you can put that here 4x minus 3 equal 10 x plus c now you can just solve for solve for y so you can see it's uh uh very simple so y equal if you know general solution is 3 minus 4x plus tan x plus c so that's the general solution and if in the initial uh if you solve the initial value problem so you can use initial condition so the initial condition y e y zero equal four gives so we plug in there so the four equal three uh minus zero plus tan c so this says
1:36:40
that tan c equal 1 in other words c equal pi over 4 that's just one value there are infinitely many values so this is one solution okay uh so so one solution c equal pi over four uh that means you can plug in that and you can write one solution as three minus four x plus 10 x plus pi over 4. if you know the general solution we know that uh for 10 you to add a k pi so that's so that's going to give you like all the possible solutions because we know that the tan
1:37:23
has the period a pi so you can add multiples of pi okay good so that's the uh solution but if you need the complete solution general solution um like with all the cases so this can be negative 4 pi plus 10 we can do x plus pi over 4 plus k pi where k is a integer so that's going to give you like all the cases because the you know pi is the period of tangent function okay so yeah so you can see that so the the basic substitute method works very well and then you could solve the problem
1:38:17
so again so after you solve for u what are you going to so this is where you solve for u so you see solve for u and then what you do use a substitution uh to solve for y good so that's the what we call the basic substitution method so let's talk about what we call the next one is the homogeneous equation so this is the second meaning of the homogeneous equation uh here homogenous means uh you can write the differential equation in as a function of y of x so that's the idea here so let's look at
1:38:52
that now so is number eight homogeneous equations
1:39:05
homogeneous uh differential equations um so what is homogeneous equation this is one definition of that a differential equation a differential equation uh that can be written as that can be written as that can be written as a as d y over d x a function of y over x so if you can write the differential equation in this form we call it is a homogeneous equation okay a differential equation that can be written as d y over d x is called a is called a homogeneous equation
1:39:59
so this is like a one differential this is like the easy definition so in this case what we can do since you can write the difference equation as a function of y of x we're going to pick that so so the solution method is so we're going to let v equal y of x so that is the substitution we're going to pick the special substitution and then after that we can write this equation as y equal v of x and then what we can do is we can take the derivative so once you take the
1:40:31
derivative what's going to happen this is y prime just use a product rule so you get v plus x v prime just by the product rule so we can use this as well so we also need that so those two are always the case with homogeneous equation no changes so we can use that and then the next step is after we do that then you can you can substitute that and then you get a differential equation in terms of v and x so solve for v first so uh so we can say rewrite
1:41:12
the equation in we and x variables and then uh solve for v first
1:41:30
and then i use then use y equal v x so simply multiply by x to solve for
1:41:43
so it's all for but so that's the kind of argument uh so uh so it's not it's not that difficult okay so um so let's look at this in a problem so as the first example let's say this one solve
1:42:10
x y d y over dx equal x squared y squared uh along with y one equals a square root two so let's try to solve this equation and you can see that this is uh this is not the basic substitution this is not separable this is not illegal factor so you can use any of those techniques to do this so what we're going to do we're going to see this term so we can divide by that term both sides so that's the starting point so the answer dividing by
1:42:55
dividing by x y so we have d y over dx which is x squared plus y squared over x y uh so if you divide x squared okay let's show the steps x y plus y squared over x y you can see x get cancelled so you're gonna get the first one become x over y y get cancelled this second will become y of x okay now you can see that this is a function of um y of x because x over y is 1 over y of x so that means everything ah looks good um because the first one you can divide by x in the top and bottom
1:43:45
so you can write this one as 1 over y of x plus the other one is y of x so that is the function of y of x now what we can do we can use the substitution now
1:43:59
so let v equal y of x then we know that y prime equal v plus v prime x so we're going to substitute uh so when we substitute uh d y over d x means y prime so this side become v plus v prime x the other side is 1 over v plus y of x is v you can see interesting part this could cancel so that term going to go away so you simply have v prime x equal 1 over v if you write this using derivative notation dv over dx x equal 1 x you can see this is a separable equation
1:44:45
so this is separable so what we can do uh we can separate the two variables so if we separate the two variables we can write this one as v dv on this side the other side one over x dx now we're gonna integrate the two sides so integrate with respect to e so it's good v squared over 2 this one 1 x means ln x plus c just add c to some one side uh so now so we solve for v now what we can do we can use substitution so use y uh y equal vx
1:45:28
uh or v equal y of x to write you can write this one as one half y over x whole thing squared here this is ln x plus c now we can uh multiply both sides by 2 x squared to isolate y so this says that y squared equal 2 x squared ln x plus c uh now what we can do is we can solve for c so let's say uh using y one equal one we have uh actually one y one equals square root two square root two we can substitute so this square root two square that's a y value x value is 1
1:46:32
and then we have ln 1 plus c but we know that the ln 1 is zero so this says that 2 equal to c says that c equal 1 so we can plug in that value back here so it is 2 x squared ln x plus 1. and now what we can do if we can take the square root but be careful we need to square root you get plus or minus now the question is which one gonna work but we know that y1 equal square root 2 that means which is positive so that means you have to pick the positive root here so it is 2 ah so we have
1:47:25
so uh yeah so we have the two there so that means uh actually c equal one yeah so let's actually forgot uh so c equal one uh so um now what we're going to do we're going to square root so you only get the positive one so it is x so we this is a 2 ln x plus 2 inside and square root so we only pick the positive one since y is greater than zero the y positive because we know the initial conditions is v positive because if you had picked the other one you get negative
1:48:02
answer so that's not going to work so but so that that means if you are given that so let's say if you had if you had y1 um equal negative square root 2 if this is given you can see that c is still 1 but you have to be the negative 1 in this case negative solution because otherwise it does not satisfy the condition because we know that when you try to solve for y you get plus or minus okay so this is this is what i'm talking about when you have y squared equal let's say 4 y
1:48:42
can be either plus or minus 2. so which one it depends on the rest of the problem so that means if you are given initial condition like y 1 equals square root 2 you want to make sure i satisfy the condition that means you have to be the negative 1 in that case so here only the positive one works okay i hope you understand the uh argument there okay now let's go to the third type of substitution so the third type is what we call the bernoulli equation
1:49:11
so let's talk about that so bernoulli's uh equations bernoulli equation uh so so what's a bernoulli equation so this is the extension of the first order linear difference equation so it is d y over dx plus p x y so we still have that part but the other side is different it is y to the n power so that's a y n term so that's the difference we have the extra term on the other side y n so there are special cases to consider so if you get this one this is a bernoulli differential equation
1:49:51
with that y to the n term but there are some special cases that we need to consider what if n equals 0 so income is 0 then you're going to get dy over dx plus px y equal qx so that is our famous one ah so this is the integrating factor method okay that's the integrating factor method ah so not very interesting i mean we know how to do that okay it's interesting but we know how to do it if n equal 1 what happened so d y over d x plus p x y equal
1:50:41
q x y so if you get that we have two options one option is we can bring that this term to the other side so if you bring this on the other side so so there are two ways to do this problem so one way we can write the differential equation like this we can bring ah that term to the other side and then what happened y is common so you can see this again first order linear so this integrating factors integrating factors now if we bring this term the other side to the left side
1:51:28
so this can be q x minus p x y you can see this is separable okay so this is separable
1:51:41
so that means we can solve these two cases uh n equals zero case is indicating factor n equal one case you can solve in two different ways either integrating factor method or uh you can use the separation of variable method so what we what we are interested in this one is we are going to consider the case we are not equal to any of those numbers okay ah so that means if and not equal to 0 or 1 we have a new case okay new situation
1:52:16
so that's the bernoulli equation so we're gonna talk about that so let's talk about that a little bit more um so if n not equal to 0 or 1 what we're going to do we can divide
1:52:37
so dividing by y to the n we can divide the equation by y to the n term divided by y to the n term if you divide the equation by y to the n term you're going to get 1 over y to the n d y over dx
1:52:56
ah and then here you're gonna get px this can be y 1 minus n because we can divide by y to the n term we already have one y term so it's beginning y to the 1 minus n on the other side you become just qx okay now we're gonna do a special substitution so the substitution is we're gonna substitute for this term that's what we do we're subject for that whole term so let v equal y 1 minus n so that is the substitution you want to remember
1:53:35
okay and that's why n cannot be equal to 1 because if an equal 1 this is just 1 so it's not going to work that's not a variable just a constant so the derivative is going to be 0. now what we're going to do uh take the derivative so taking
1:53:52
derivatives
1:53:57
so taking derivatives we can write this one as uh v prime so we can just chain rule here it is 1 minus n y you subtract one more so minus n by n y prime by the chain rule so this is the chain rule
1:54:18
good uh so that means we can solve for y to the n y prime term so this says that y to the minus n y prime is simply 1 over 1 minus n times v prime so that's always the case now what we're going to do we're going to substitute that in the equation so if you go back and substitute in this form uh so you're going to get 1 over 1 minus n v prime that's the first term plus p x v equal q x and you can see is again first order linear in v so this is first order linear
1:55:15
first of the linear in v that means we can use integrating factors okay so i use integrating factors
1:55:30
use indicating factors to solve for v and then i use v equal y to the 1 minus n to find y so that is the argument okay so that's how you solve the bernoulli equation so let's look at uh one example to see how it works get um so how about this one solve
1:56:09
let's say uh 2x d y over dx minus y equal 6 x squared y cube this one ah so you can see we have this term and this is bernoulli it's a bernoulli equation so we divide by that term and also you can see that there's a unnecessary term here
1:56:45
so what's the term we have here is and you can see
1:56:55
so uh dividing by um we also gonna divide by x also uh actually you can divide by two also but we can leave two and you can see how it works so um so dividing by
1:57:14
dividing by x y cube so you're going to get 2 y cubed by negative 3 d y over dx minus 1 x y negative 2 equal one of the external cancer y3 get cancels uh so you're gonna get c uh 6x so that's what you get now the substitution so we divide now you can see that you get to the form that we want now we're gonna substitute for that ah so we're gonna say v equal y negative two
1:58:04
uh this implies that we prime equal negative 2 y negative 3 y prime in other words y negative 3 y prime is negative one half v prime i mean that's what happened so we read like that and now we can substitute that so if you substitute that so by substituting
1:58:33
substituting 2 times negative 1 half v prime substitute for that and then minus one half v equals 6 x you can see this get cancelled and uh and then you're going to get negative v prime negative 1 over x v equals 6 x you can multiply by the negative sign so v prime plus 1 of x v equals 6 you can see this is first order linear and v that's exactly what we want first order linear in v so let's use the integral factor method so the integral factor uh integrating a factor i equal e to the p x dx
1:59:39
so e to the 1 over x dx e to the ln x that means x
1:59:49
so we get that now what we can do we're going to multiply by that both sides
1:59:56
so multiplying by x
2:00:04
so we're going to get x v prime minus and plus v equal negative 6 x squared this side is automatically the v and the integrating factor which was x equal negative 6 x squared now integrating those steps are just the uh integrating factor argument integrating we're going to get v x equal if you integrate this one it's going to be negative ah so you're going to divide by 3 so it is 2 x cubed plus c so that means v equal if you divide by x is negative 2 x squared plus c
2:00:51
over x now what you can do ah using ah v equal y to the negative 2 we're going to substitute that so we have 1 over y squared which is y to the negative 2 equal negative 2 x squared plus c over x so what you want to do now is you want to flip it over but before you flip you're going to get the common denominator here you have to do that so x uh so that means you're going to multiply by x what is you can write this in a c minus 2 x cubed now you're going to flip it over so that means y squared is
2:01:33
x over c minus 2 x cubed now again considering our cases you can say y equal since we don't know we get i plus or minus x over c minus 2 x cubed square root so those are all the cases the two cases and if you know the initial conditions we can find them otherwise this is the general solution so this is the general solution okay so that's about that so that's the basic uh that's the substitution method so we talk about the basic substitution uh homogeneous equations and then the
2:02:18
bernoulli equations so and then the uh the last type of first order differential equations the exact differential equations let's talk about that now uh so this is the last type we talk about what we call the exact uh differential equations
2:02:40
so what we mean by exact difference equation so we can look at the standard form which is m a function of x y d x plus n a function of x y d y equals 0 so that is the standard form we're going to consider okay it has to be dx plus d y ah but there are other ways to write the same equation for example if you divide the equation by
2:03:12
dx uh you can write this same equation this form as well so you could write this one as uh for example d y yeah i'm gonna write it uh down here because i need some space so we can by dividing by dx we can write this one in for example d y over dx times times n x y and xy plus mx just by dividing by dx but we don't do that because this is really not the right way to do it uh we're going to go with the the standard form above so this is what we call the standard form
2:04:00
we do that for a reason because there are different equations we can come like in physics and mathematical physics come into this form so that's why we kind of study that now let's talk about the uh the theories uh that we discuss so what we call if the da ba m there's a partial derivative of m with respect to y equal the the partial derivative of n with respect to x if those two are equal we call this differential equation exact equation okay that's the kind of test uh the equation equation
2:04:39
is called exact is called exact differential equation uh this is the kind of test we use uh in that case and in that case what happened there is a function there is a function there is a function if x y equal a constant so there's a function like that uh such that such that the partial derivative with respect to x equal the m function and the partial derivative with respect to a y equal in function so it says that if that is the case if this condition satisfies this condition
2:05:33
if this condition satisfies there is a function such that f x y equal constant c is a constant uh that can give you double f over double x equal m and double flow down by equaling okay and then if you know that function f x y what we're going to do we're going to solve that function we're going to solve for y so that's what we do and then that y is the solution we are looking for so let's talk about let's look at a problem um before we do that actually let's talk
2:06:06
about the solution method so what we normally do is we're gonna start with one of those equations there are two equations so we start with the easy one and then take the integral once you take the integral then you get the function back so that's the kind of idea so let's try the steps here so let's call the solution method
2:06:28
solution method so this is the lengthiest argument that we use in differential equation ah yeah so let's say you can pick just one of them so let's say you uh go with the x function uh x derivative so let's say you pick that one so you can just reverse the same argument you can just rewrite the argument using the other one if you want but let's go with this and then what you're going to do since this is a partial derivative with respect to x we
2:07:00
can integrate with respect to x so it's very important so integrate ah with respect to x so if you do that you get the function back it's a function of x y this integral of m x y with respect to x since this is a partial integral maybe you had a function with only a function of y because since you are taking the partial derivative with respect to x to find double f over delta x maybe there was a function of y alone so we just put that back in so that's a new term it's not a constant it's a
2:07:39
function of one now what we're going to do you're going to combine the two so we're going to differentiate this one so differentiate with respect to y variable so if you do that then you're going to get da by four over double y that means you can integrate whatever the answer you get uh with respect to y because a function of x and y we want to integrate with respect to x so what will happen here it just becomes this regular derivative because the
2:08:09
function of y only we don't partially do this there now what we can do we can solve for g y but the goal is to find this that's the only thing we don't know so the goal is find okay that's the goal find this so now what we can do we know that this should equals the uh the inverse function because this is the partial data with respect to y this is exactly the n x y function so we're gonna set it here we're gonna set that we're gonna say this equal to this now what we're gonna do
2:08:43
solve for g prime y okay or simply solve for g prime y because you can find it because because they match then you can find g prime y and then ah integrate and find g y so the next thing is uh first find g prime y and then uh solve for
2:09:13
or integrate and then we can integrate that first find g y and or let's let's write like that first find g y and
2:09:27
and integrate it's a derivative integrate to find g y now what we're gonna do we can substitute that in the uh just the function above once you substitute that there so what we can do we can substitute the substituting substituting we have f x y function so we substitute here we know this part plus now we know g y because we found that above okay now we are not done yet but we know that the function what we can do we're going to set this equal to we can say this equal to
2:10:15
a constant because that's the function we are looking for we know that there was a function which is f x y is equal to cons to a constant so we sit there and then solve for y if possible after that what we're gonna do solve for y if possible because sometimes uh we cannot do that uh but if you cannot do it you get a implicit solution for y so you get implicit uh solution for y but if you cancel it's the implicit solution so that is the solution method i know it's little lengthy
2:10:59
uh but that's how it works and then we also talk about a shortcut method um so i'm gonna discuss that so so you can use that if you remember uh that's that's not that is difficult it's very easy so let's talk about shortcut method uh so the shortcut method ah so what was that so i'm going to write that you may use
2:11:30
you may also
2:11:34
you may also use the shortcut method
2:11:46
uh so what's the road card method uh we can start with f over delta x equal m x y function
2:11:58
and c is the partial derivative with respect to x what we can do we can integrate this one with respect to x so integrate ah with respect to x once you do that you get the function so what happen when you do that i ah some space here once you do that uh you get a func you get the function f x y
2:12:31
so you get a function you just integrate with respect to dx so when you integrate you can see that you're going to see simply consider y is a constant so you get some expression and do the same thing so what we do by 4 double y which is n x y what we can do with this you integrate with respect to y so you again get the same function in another form so here what we did we did the partial integral with respect to y now what we can do we can write the common terms
2:13:12
once uncommon terms once so your function f is simply you write common terms common terms and uncommon terms
2:13:38
once so that's the uh again so after you get that now what we can do as before uh we're gonna set f x y to a constant and then uh after we do that so we sit after that what we can do uh solve for y if for simple okay so that's the that's a shortcut um so you start with double for double x integrate with respect to x you go with double for double y integrate itself to y and you get two expressions you write the common terms once right the common terms you can see some
2:14:22
common term ones an uncommon term once that's the function f x y you say this equals zero solve for y is possible so let's look at a problem uh so this is so let's say solve 3 e x minus y squared dx plus 3 minus 2 x y plus sine y i d y equals 0. so if you get a problem like that the first thing that we do is we check whether this is exact so this is the m function this is the n function so what we're going to do uh we're going to check we're going to do the test here so the
2:15:24
test is we can do da by m over double y remember the m function do with respect to the other one so double y so which is negative 2 y and you can see that that's exactly equals to da by n over delta x ah if you respect to x the constant sign term goes away with this partial derivative you get negative 2 y so they are equal that means it's a exact equation
2:15:54
okay that's an exact equation so that means um
2:16:00
so so that's that means that so there is there is a function
2:16:11
there is a function z if there is a function z which is simply f x y equals c such that so there's a function z which is a function of x y such that by f over delta x equal the m function which is 3 e x minus y now what we can do we integrate respect to x
2:16:52
so integrate uh with respect to x so if you do that you get f x y if you integrate with respect to x it is 3 e x minus x is y minus x y squared plus g y because we don't know whether there is a function so i'm going to call this equation number one now what we're going to do we can differentiate this one differentiate with respect to the other variable y so if you do that you get da by f over double y if you differentiate with respect to y this is 0 minus 2 x y because it's a y so you get 2 y
2:17:41
it minus x you get minus 2 x y and then plus g prime y because a function of y only so it's just the regular derivative now we're going to set this equal to the n x y function because we know that by the theory there is the n x y function we can set it uh yes it to that ah so it is 3 minus 2 x y plus sine y so is clearly you can see that this function has to be 3 plus sine y when you set them equal g prime y has to be 3 minus sine y because they are equal to each other good okay so we know that
2:18:35
so that means so this implies that g prime y equals to 3 plus sine right now what we're going to do we're going to integrate with respect to y so integrate ah with respect to y and if you do that you're gonna get g y equals to the integral of g prime y d y uh that means if you integrate it's a simple uh integral with respect to y so you can see that you're gonna get 3y plus this is sine so sine means negative cosine negative cosine y and we are not adding plus c here
2:19:29
now by equation one since we find y we can substitute by equation 1 if x y equal 3 e x 3 e x minus y squared x plus 3y minus cosine y so that is the y g y function that is g y so what we do are we going to say this equal to c by theory so that's a constant function f x y that satisfies those conditions so now what we can see there's no way you can solve for y so then the implicit solution is the implicit uh solution is given by 3 e x minus y squared x plus 3 y minus cosine
2:20:29
y equal constant so that is the implicit solution for y and you can see that uh since we have the cosine y term and the y squared there's no way you can solve this for y but if you do not have cosine y term then you can solve this for y because you can use the quadratic formula um and then you can solve for y but since you have the cosine y term we can solve good so this is how this is how we solve an exact differential equation so next we can talk about what if
2:21:03
it is not a exact equation so what we call the uh almost exact equation so we're going to solve that next so let's next talk about how to um solve almost exact equations so almost exact
2:21:29
so what's going to happen uh if you look at the standard form of the exact equation m x y d x plus n x y d y equals zero so what can happen when you take the partial derivative the first one with respect to y it does not equal the partial derivative with this of n with respect to x does happen so in this case you can see that the equation is not exact ah it's not exact so what we're going to do we're going to use the integrating factor method so we can use the integrating factor method
2:22:18
so we use integrity factor method so there are two cases to consider uh we have two cases
2:22:28
so um so the first one is we're gonna calculate my minus n x over n we calculate that if this is a function of x only if this is a function of x only so let's assume that this is a function of x only it's a function of x only in that case we're going to calculate the integrating factor is also going to be a function of x ah which is e to the we can integrate that expression m y minus n x over n with respect to x so that's the integrative factor in that case the other case is ah nx minus m y over
2:23:28
m so this is a function of y only if this is a function of y only then you're going to calculate integrating factor i y which is e to the which is e to the um n x minus m y over m t y now so how you can remember them um so you can remember them like this so you can see this is um uh m n n the other one is n m m so that's the format of that m n n or n m m um good and then after we find the degree factor the the the rest of the uh problem is exactly the same as what we
2:24:27
normally solve a problem in uh using integrating factor methods after that what we can do then we're going to multiply by the integrating factor then we're going to get i times m dx plus i times n d y so this is the exact equation so we're gonna follow the exact uh equation argument this is an exact equation uh so we're gonna follow the exact equation argument so let's do just to one example just to see how it works uh good so let's solve this problem so let's say solve x y d x plus 2 x squared
2:25:14
plus 3 y squared minus 1 d y equals 0. so let's solve that so what we are going to do we are going to pick this as m we are going to pick this one as n uh then you can see that da ba m over double y which is x and by n over double x
2:25:47
uh which is 4x and you can see that they are not equal so this says that by m over double y not equal to n over delta x so that means not exact so not exact all right so what is going to be gonna see uh which one gonna work so let's calculate uh my minus nx over n first so what do you get what is m y what is the y derivative with respect to y derivative of m which is x and a minus let's calculate the x derivative of n which is 4 x divided by n n means 2 x squared plus
2:26:44
3 y square minus 1 which is ah you can see that ah this is this is not only a function of one variable ok so you can see that not a function of one variable so not a function of one variable
2:27:13
so that means it's not going to work so let's look at other one uh so in x minus m y over m so if you do that you're going to get nx that is 4x and y that is x divided by m now m is x y so if you cancel x so it's going to be 3x over 3 over y so you can see that this is a function of um y only so the function of y only so that means we can calculate the you can find the negative factor so i y e to the uh the formula n x minus m y over m d y we already calculate that so it is 3 over y d
2:28:15
y that means 3 l n y so bring it here so that means going to be y cube so y cube is the um indicating factor so i'm going to multiply by that
2:28:36
so multiply by y cube um the differential equation so you're going to get x y to the fourth power dx plus ah you're going to multiply by y cube so it is 2 x squared y cube plus 3 y to the fifth power minus y cubed d y equals zero so we get and you can see that once you do that it's going to be exact with us how we pick the numbers so now this time we can use the shortcut let's use the shortcut to get the answer so this is exact ah so we use the shortcut method so we use
2:29:28
uh the shortcut method okay so let's use the shortcut method so if you the shortcut method uh what do you get uh we're going to go down by 4 by x which is the first function so we're going to integrate this one with respect to x so if you do that you get f x y equals to x squared over 2 y to the fourth power ah will be integrated with respect to x and we are not going to add the constants and then go with the other one by f over double y which is the other function uh
2:30:19
so it is 2 x squared y cube plus 3 y to the fifth power minus y cube uh we can integrate uh this one with respect to y if you do that you get f x y uh equal one half x squared by 4 plus 1 half y to the 6th minus 1 4 y to the fourth power now what we can do your f x y function is we're gonna first write the common terms and you can see this is common so we can write that one so one half x squared y to the fourth power and they write the uncommon terms so one half
2:31:20
y to the sixth minus one fourth y to the fourth now what we can do we set this equal to zero and you can see that uh it is not easy to solve this for uh y so that means this is the implicit solution uh this is the uh implicit
2:31:45
implicit uh equation implicit solution in the equation for what so that is the that is how we solve the uh almost exact equation and there are like you know cases uh you cannot really solve uh because we only consider two cases uh whether my oven x over n e is a function of x or in x minus m y over m is a function of y what if it is not the case so that means there are some cases that still give you uh problems okay and then there's a video um what we normally call it
2:32:26
all in one video it's already posted uh so so that's the kind of good one that's gonna give you kind of all the methods in one video so you can look at that you see i'm gonna also put the link to that video all in one review uh all in one review for first order equations okay so it's gonna show you all the uh formulas there good uh now what we're gonna do um we're gonna uh discuss uh the numerical solutions and you can see that there are a lot of equations
2:33:04
uh that kind cannot solve okay there are only few equations just a very small percentage of equation that you can solve so that means there's a need to solve uh we need to find approximate solutions to find approximate solutions uh we're going to use numerical techniques so let's talk about that so that is number 12 are numerical methods
2:33:27
uh numerical methods
2:33:33
in vegeta called numerical methods uh there are actually discussed three of them uh one is the easiest one the most obvious one what we call the euler's method ah and then we have improved euler
2:33:55
improved euler's method and then the very good one what we call the runge-kutta method
2:34:09
so the runge-kutta is like very good it's going to give you answers like um excellent answers for certain problems so let's talk about the euler's method that's the easiest one
2:34:27
euler's method uh so how it works so what's a geometric uh argument for that so let's say you have a curve ah so what's gonna happen you're gonna you are given uh a point and the derivative like the slopes everywhere so you are given the derivative at each point and a starting point so that's the only information here you can find derivative at any point and also give you the starting point so the goal is to find an approximation for another point at some point
2:35:05
so that means um uh so so so this is the actual curve so what you trying to do you're going to approximate the value at a certain point so what we normally do is we're going to divide this into equal parts uh and then calculate approximate the value successively so you're going to calculate y1 ah that's y1 and then you calculate y2
2:35:34
and they need to calculate y3 like that so uh so that's the kind of argument we do we're gonna approximate uh one by one uh we can look at the tangent line that's how we do we can look at the tangent line and then calculate uh the next one uh so actually uh the first one is not exactly correct so um because when you draw the tangent line uh so what happen is let's extend the tangent line so this is what you get so that means your first value is here
2:36:06
right here that's the first value so that's the first approximation so that is y 1 now we can use that value and also use the slope at the point to find the next value so if you draw the slope at that point you can see it's going to go like that so that is y2 that is y2 so like that we're going to find that's y2 and then we can draw the slope at that point so it's going to go right here that's because we're going to parallel to that this is slope so it's going to give you uh y 3 so that's y
2:36:48
3. so that's the approximate values of y 3. so this is the kind of idea so you're going to look at the slopes at each point and then use the tangent line to approximate the next point so that's the kind of idea uh so what we normally do is we're going to divide this into equal sub intervals h h like that and every time calculate the next point calculate the next point so what you are given is simply you are given dy by dx fxy so that means you know the slope at
2:37:29
every point and also you know the starting point so those are the only things you are given the slope uh at every point and the starting point
2:37:45
so ah and then ah our formula becomes uh like this so your n plus one node is simply x n plus the gap and then your y n plus one the next one is simply uh y current one plus h times x n y so that is the scheme for the euler's method so this is the wheeler's method formula and this can work you can start with n equals 0 once you plug in 0 you get x1 and you plug in 1 you get x 2 3 like that depending on what how many points you need okay so that's that's it so that's
2:38:33
earliest method let's just look at an example um and see how it works so let's look at ah example so let's do this example let's say approximate
2:38:52
approximate um y2 using euler's method
2:39:07
using wireless method
2:39:13
if given that
2:39:19
if given the differential equation d y over dx equals say 3y minus x and y 0 equal minus 1 and the gap the step size 0.5 so it's a kind of problem that you can get you can differential equation initial condition and a h so uh it is always a good idea to kind of uh try to understand what's going on here so we have x zero the starting value that is uh zero and then we have y zero that is uh the y value so how many we need to calculate so you can see that
2:40:02
x zero is zero and then we're going to go in 0.5 steps so that means we have 0.5 and then 1 1.5 and 2. so you calculate at 2 so that means x 1 x 2 x 3 and x 4. so so knowing x naught and y naught you want to calculate y 1 y 2 y 3 and y 4 and calculate ah formal so let's look at the uh situation here so uh so what we have is so uh so the answer so what we can do we're gonna first identify what is the x uh y function so you can see it is 3y minus x that's the xor function now we can add
2:40:54
the scheme so what's the scheme x n plus 1 is x n plus h so in this case it is uh x n plus 0.5 so every time you increase by 0.5 and then y n plus 1 which is y n plus h f x n y n so let's plug in and simplify so this is y n plus h is 0.5 the function is 3 y n minus x n so you plug into the function that's a function so plug in for that now what we can do is we can simplify if you simplify this what do you get you're going to get 0.5 times 3 means 1.5 plus 1 so it is 2.5 y
2:41:49
n minus 0.5 x n so that is the formula that we're going to use for each n so we can start with we have x naught which is 0 and we have y naught which is negative 1 now we're going to start plugging the value starting first value is 0 because when you plug in 0 you get x 1. so when you plug in 0 you get x 1 which is you add 0.5 so it is 0.5 and y 1 the formula says it is 2.5 y naught minus 0.5 uh x naught since you know the two values we can plug in so it is
2:42:34
point uh 2.5 y naught is negative one and then yeah y naught is negative 1 and then we have 0.5 times x naught is 0 so if you simplify so you can use a calculator for that you're going to get negative 2.5 so that is the y one value so keep doing so now you plug in n equal one if you by n equal one you get x two which is a one point 1.0 we have y2 now so it is 2.5 y1 minus 0.5 x1 since we know uh y1 and x1 we can plug that's coming from the previous step it's always those two coming from the
2:43:25
previous step so we're going to plug in those two so you get 2.5 y 1 means negative 2.5 minus 0.5 x 1 means 0.5 always kind of on the previous step so if you use a calculator you get negative 6.5 go to the next one n equal 2 so this is going to give you x 3 which is 1.5 is going to give you y 3 value according the formula it is 2.5 by 2 minus 0.5 x 2 we know all the values from the previous step so it is 2.15 negative 6.5 minus 0.5 and then
2:44:25
x2 is 1.0 now what we're gonna do we're going to simplify again so you're going to get negative 16 7 5 is a calculator and then we can by n equal 3 you get x 4 that means that's exactly the number we want y 4 that's the value we are looking for again it's the same formula it is 2.5 by 3 we know that and the 0.5 x 3 we also know that so let's plug in there so you get 2.5 negative 16 7 5 minus 0.5 and then 1.5 so if it's a calculator are you gonna get negative 42
2:45:16
6 2 5 but this method is not very accurate so you don't need this many decimal point we normally leave like just one decimal point so that means your y two value which is the y fourth y two minimum means the y four uh is approximately equal to negative forty two point six okay so you can see it's not that difficult um so it is always a good idea to uh get this first uh because then you can simply plug in there uh every time rather than using the original equation every time so always
2:45:55
try to simplify that first and then use that simple equation this is much much uh easier okay so ah so this is how you use the euler's method to solve find the approximate solution so next what we're going to do is we're going to discuss the most uh correct version what we call the runge-kutta runge-kutta method ah so this is very very accurate even the commercial softwares use that so i should say that so this is a excellent result um uh like a very similar problem so you
2:46:41
have y prime equal f x y so the slopes are given with a initial condition y x naught equal y naught so you have a uh so you know the starting point and you also know the slope so then the goal is to approximate the actual curve the same idea uh we're gonna uh split uh as before though we're gonna call the gap is h and uh into approx find the value at x n so uh so the scheme is slightly different now so what we're gonna do we're gonna calculate several intermediate values so we're gonna go for n
2:47:29
equal zero one two like that depending how many you need as before we're going to calculate x naught which is x n plus 1 equal x n plus h no change but before we calculate why we're going to calculate several other slopes so we can calculate k 1 slope that is the slope at x n y n then we're going to calculate k 2 slope uh that is a slope uh halfway in between so you know x in you go h over 2 comma y n plus h over 2 times k 1 and then k 2's k 3 slope which is f
2:48:15
again at the halfway point comma y in plus h over 2 k 2 now so we have k1 k1 goes there and then k2 goes there and then we have k4 so what is k4 which is f x n plus h y n plus h times k 3. so we're going to calculate those extra slopes and after you calculate the slopes then your y n plus 1 the next approximation value is given by y n plus h divided by 6 um k 1 plus 2 k 2 plus 2 k 3 plus k 4 so that is the runge-kutta scheme is very similar you are you know x naught y naught so you're going to
2:49:10
calculate x 1 y 1 x 2 y 2 x 3 y 3 like that so this is what we normally call the fourth order fourth order so normally call rk with it method uh these are normally uh used in what we call ode45 algorithm in matlab and other softwares are used and improved use an improved version use an improved improved a version of this
2:50:01
of this okay so so that means it's very accurate the solutions are very accurate and we solve several problems in the homework uh and look you can look at that uh also we have solved several problems in the uh video as well so if you need more information you can find them uh good so this is the runge-kutta method and i'm not gonna solve the problem from this one because we're very lengthy uh because you know that there are a few many things you need to calculate before we calculate the next one
2:50:33
okay so this is how we gonna solve a uh this is the next good numerical method to solve a first order differential equation next let's talk about uh what we call the directional fields directional
2:51:02
fields some also call this slope field slope field uh so what's the directional field so what we're going to do uh so when we have a differential equation so let's say you are given a differential equation d y over d x equal f x y so what we can do is at each point at each point in the domain
2:51:42
draw an arrow draw an arrow indicating the direction of the tangent line indicating
2:51:55
indicating the direction of the tangent line the direction of the tangent line tangent line so that's what we call the directional field uh or the slope the direction of the line means the slope so we can do a look at a problem so let's say
2:52:31
i draw
2:52:35
the directional field field of let's say rdy over dx equal to x minus y so if you get a problem like that uh what we're gonna do we're gonna draw like a typical grid uh you know with enough points um so let's do that so this is a grid
2:53:15
with enough number of points
2:53:23
so normally like this will be given
2:53:32
so let's divide this into three parts
2:53:48
so what we're going to do let's say this is 0 and then 1 2 3 and then negative 1 negative 2 negative 3 like that
2:54:03
so this is 1 2 and then negative 1 negative 2 so what we do is so this is x direction and this is the y direction so we're gonna at each point you can evaluate the slope and draw a small line and we know uh if there's a positive slope so the error goes like that this is for positive slopes if you have a negative slopes thus the error goes down this is zero slope is horizontal so we can use those three facts uh so let's evaluate at zero so at zero you can see that if you plug in zero to
2:54:43
the differential equation you get d y u by dx six zero so that means that's at zero slope is zero so you draw a line like that with zero slope and then also you notice that there's a special line for example if you look at this line uh so you can say on the line on the line uh if you set this equal to 0 so you're going to get y equal 2x on the line y equal 2x the slope is zero the slope the slope is zero
2:55:25
slope is zero so that means what is this line uh so that means at uh x equal 1 y equal to so that means this is the line so the line uh passing through these points so on this line the slope is zero so that means if you pick any point on that line you know that slope has to be zero so that's one thing so we know that on that line slope is zero and also let's calculate few other values so for example at this point if you look at this point
2:56:03
you can see that it is x equal 1 y equals 0 so the slope is 2 so it's a line with slope 2 and at that point you can see um slope is four so it's more steep and here like slope is six so it's much more steep like that and if you look at the other side uh you can see at negative one slope is negative two and negative 4 and negative 6 is much more steep and then if you go along the y axis you can see that x coordinate is zero y coordinates increases but
2:56:40
with the negative sign uh it's going down so for example at this point the slope is negative 1 so it's going to go like that with slope is negative 1 and if you go up it's negative 2 so it's much more steep and here at negative 1 it is 1 and this is a 2 and we can also draw a few more so for example if you add 1 1 so you're gonna get slope uh 2 minus 1 so it's 1. so the slope is exactly one here and then if you go further it's the same y coordinate but x squared increases so we're going to get
2:57:21
more steeper like that and then here what happened x is negative 1 that means negative 2 and then y is uh 1 so so at this point what's going to happen is uh let's calculate what you get so you can see that x equal negative 1 that means it's negative 2 and 1 is negative 3 so it's negative and when you go up it's getting more steeper because y code y is sub you subtract what that means get more steep and at negative two it's given a negative four and so it gets more steeper you go
2:58:02
the same thing when you go this way it's gonna get much more steep and uh what about this point so this point uh you can see x is one so is two and then uh minus one says three because y is negative so it's three so it's more steeper the same thing much more steeper when you go that way ah so go down that means gonna get more steeper like that so we only left these so let's calculate what happened at this point so which is negative 1 that means negative 2 and then uh and then we have uh
2:58:43
y equals negative 1 the negative has become positive so it becomes negative one and then on when you go it's going to get much more negative like that so those are a few uh values okay so we we still have just a few more uh so at this point uh what's gonna happen is it is negative two that'll be negative four and with uh negative uh two is become uh so what happened here so you can see when you go down uh you subtract negative values that means can i get less deeper so like that uh
2:59:19
yeah because because the negative sign so it's still less deeper when you go down so that means this is this has to be less deeper than that good so when you go down it's getting less deeper because the negative voicing okay so that's the uh directional field uh for this one and
2:59:46
good so now what we can do is we can draw a typical curve like for example if you want to draw a curve that passes through 1 negative one one so that means the point at this point that's the negative one uh one point so what's gonna happen it's gonna follow the uh arrows so so if you follow the arrows what's gonna happen it's gonna go like that go like this and i think it's gonna go like that and get zero and then turn and i'm gonna go uh like this
3:00:20
so this is a typical curve that passes through uh the point one one so what we're gonna do is simply draw follow the uh arrows so like that so good okay so that's what we can do with uh directional fields and then uh what else we had uh uh we talked about uh existence and weakness theorem so let's talk about that a little so we have the existence and the uniqueness theorem number 14
3:01:02
existence and the uniqueness
3:01:11
existence and uniqueness so let's look at the differential equation d y over dx equal f x y along with y x naught equal y naught ah so let's say so there's a region so there's a point x naught y naught so so we know that point now the question is whether there is a unique solution that passes through that point actually uh that depends on certain conditions of the function uh so for example if the if the function f is continuous and it's y derivative
3:01:52
why partial derivative is continuous so the answer to that question is yes there's a unique solution so let's try that as a result uh so but in general uh a solution a solution may or may not exist or may not exist but under certain conditions uh the answer is yes so let's uh write the theorem so this is what we normally call the picard's existence and weakness theorem
3:02:33
so because
3:02:40
existence and uniqueness there
3:02:48
so what it says that for the initial value problem potentially problem d y over dx equal f x y along with y x naught equal y naught there is a unique solution there is a unique solution this unique solution
3:03:21
just unique solution near x naught y naught if the function f x y and its partial derivative with respect to y are continuous
3:03:46
near x naught y naught so we check what happened near x one why not if the two functions f x y and the d by f of the double by the positive with respect to y continuous then there is a unique solution around that so that means if you can extend that interval along that interval uh there is a solution so it can be the whole region because this this interval can be the whole domain so that's the uh exercise theorem so we only check whether these two functions are continuous
3:04:19
near that point so let's just do one example just to see how we can apply that
3:04:28
so for example something like this for the ivp for the initial value problem rdy over dx equal x over y along with y1 equal 2 so that means you can see your region is so say 0 so it's a 1 and 2. so there's a point somewhere so we're going to check what's going to happen near that point in that box you can see there's a box defined
3:05:12
yeah okay so we can check whether there's a solution in that small box this box can be everything but we just guarantee in that small one okay so then we can see that there is a there is a unique solution because you need a solution
3:05:39
y equal yx as a function of x as it means near because it near
3:05:51
the point near the point 1 2 since the function f x y which is x over y and double f over double y if you take the positive with respect to y you're going to get negative x over y square that's what you get you can see that though both functions are continuous there are no problems with that near that region but you cannot include 0 if you include 0 we get the division by 0 situation but here no because we only consider near one two points so everything ah all the funds are continuous so since
3:06:34
the two functions are
3:06:40
continuous continuous near the point one two
3:06:54
uh but there are problems uh near uh you know one near uh one zero if we have one zero we got that point the y doesn't exist anymore okay so
3:07:17
so we can see but near 1 0 the solution
3:07:30
does not exist
3:07:50
by
3:07:59
f x y equal y of x discontinuous
3:08:13
this continuous at once here that means there are no solutions for that point there's no slope at that point okay so and then uh uh we need to talk about the autonomous difference equations the autonomous differential equations
3:08:38
autonomous differential equations so you can see a differential equation insurance liquidation is called is called autonomous
3:08:58
it's called autonomous uh if it does not depend on the independent variable if it does not depend on the depend on the independent variable
3:09:22
independent variable
3:09:27
for example dx over dt equal minus x x minus 1 squared is autonomous because you can see that it does not change over time because there's no t term so it's just fixed okay so this represent the uh what we normally call that stable solutions stable situations because it doesn't change difference equation doesn't change it can stay there so so you can see that so the knot so you can see does not change does not change does not change as does not change as time that means in this case t changes
3:10:32
okay that's very important uh and then we define what we call critical points for that so critical points so critical point means uh the places where the derivative is zero okay so if dx over dt equals 0 at x equal x naught x naught is called called a critical point
3:11:11
it's called a critical point and then after that what we did was we check whether the critical points are stable unstable half stable so uh so we may check we may check whether
3:11:29
the critical points critical points are stable let me check whether the critical points are stable
3:11:43
or not so let's do that in a problem so we can do uh something like this so the example uh find the critical points and find the critical points
3:12:09
find the critical points and classify classify them as stable
3:12:26
half stable
3:12:31
or unstable yeah so there are three types stable has stable ones tables ah for the differential equation dx over dt equal negative x squared x plus 1 x minus 2 something like that so what we can do we're going to first find the critical points so for critical points for critical points what we can do we're going to say dx over dt which is negative x squared x plus one x minus two equals zero this implies that from each factor the first factor say zero um and then we have uh negative one and two
3:13:32
so negative one from here negative one coming from here and the two come from there so those are the three critical points now what we're gonna do uh we're gonna draw the uh phase portrait so it's a vertical line and we're gonna put those numbers in the order so we're gonna go with negative two first and then we have zero and then we have two so those are the three points that we have in that order now what we can do we're gonna pick uh
3:14:01
points in those intervals so we're gonna pick a point which is greater than two we're gonna pick a point between zero and two and then pick a point between negative one and two negative one and zero and then pick a point which is less than negative one now what we're gonna do we're gonna check the sign of the slope uh so we can take the sign of the slope so you can see that uh you just plug in a just point uh in those region to the equation and you can see that uh it is
3:14:35
uh if you pick a point like let's say three so a negative and if you plug in three it is positive and if you plug in three it is uh four is positive if you plug in three uh is positive so that means this says that uh it's a negative positive positive positive so it's negative that means dx over dt is negative that means it's gonna go down so we're gonna put the arrow going down do the same thing uh now we're gonna pick a value between 0 and 2 so like for example 1
3:15:07
so if you plug in 1 you can see it's negative positive this automatic and then if you plug in 1 you can get x plus 1 means 2 so it's positive and if you plug in 1 this is negative so that means it's going to give a negative negative a positive sign so that means the x over dt are positive so that means it's gonna increase so we're gonna draw a line like that at all like that now we're going to pick a value between 0 negative 1 like a negative 0.5 so if you do that it's negative positive
3:15:40
automatic and if you pick a value between uh like as like a 0.5 and then you can see this is positive and then this is negative so that means positive so that means dx over dt the this is the slope uh which is positive so that means it's going to go up so that means the error going up now for the last one we can pick like something like negative 2 so if you pick negative 2 is a negative positive if you pick negative 2 is going to be negative this is negative
3:16:14
so there are three negative signs that means negative so that means dx over dt are negative so it's going to go down so we're going to put a down arrow so now you can see that there are the two arrows converge in the first one so that means this is a stable point at this point is stable i mean both sides gonna converge to that and at zero you can see that one goes in one goes out so it is half stable half stable and then the last one you
3:16:45
can see that going outside is unstable so we can see that for those critical forces that's the classification and the last thing what we can do let's say um you start with the point like at the beginning you start with a point so let's say so the question is if x 0 the starting value is always say negative 0.5 so what's going to happen in the long run so find uh the limit find the limit t goes to infinity x t so what's going to happen you can see that
3:17:25
when time goes on it's gonna go up so that means uh it's gonna converge to zero but even though the all the arrows uh the next arrow is is also going up it never go to two because it's never passed a critical point that's a very important fact name across a critical point so never cross a critical point
3:17:58
okay so this is all about first order differential equation and related topics and then we can talk about second order differential equations
3:18:10
so this is the final exam uh review of part two in this uh we are going to talk about introduction to second order linear ordinary differential equations in that we can talk about what we mean by general solution and the linear independence including wranskian test then we can talk about the reduction of order method uh that means we can reduce a second order differential equation to a first order differential equation if you know one solution and then we talk about constant
3:18:40
coefficient second order differential equation then we talk about one of the very special case what we call the kochi euler difference equation then we talk about the higher order constant coefficient differential equations and how to solve them using what we call the characteristic equation method and then we talk about the non homogeneous equations under that we're going to talk about undetermined coefficient method and the variation of parameters method so uh
3:19:12
let's start with the second order linear ordinary differential equations so this is number 16 according to a list so this is second order linear second order linear ordinary differential equations
3:19:40
ah so let a y be the dependent variable dependent variable and x be the independent variable
3:20:07
it's always a good idea to check what is the independent variable um because sometimes we're going to confuse and write the wrong one and everything is wrong after that so if you look at a general second order linear differential equation you can write that in this form so you can write this as ax that's the coefficient function of the first uh the highest term and then we have bx there is another coefficient function of the second one and then plus c x y equals dx
3:20:37
because x is the independent variable so you can write it like that that is the general second order linear differential equation now what we're going to do we're going to divide by the ax term just to make it easy and to bring it to the standard form so we can divide by x so dividing by x dividing by ax we can write this equation as y double prime plus bx of ax bx over ax later equal the p x and then y prime plus c x over ax y equal a dx over x so we can write like that now what
3:21:23
we're going to do uh we're going to call this uh bx of ax px and this is qx and this is our fx so with that we can write the standard form so we have the standard form standard form y double prime plus p x y prime plus q x y equal f x so this is what you get and this is what we normally call a standard form um so if the because we know that ax is not zero completely zero so therefore we can always do that it can be zero what we can do we can remove those points so what we normally
3:22:08
call the singular point so we remove them uh now what we know this is second order differential equation so there should be uh so this is second order uh ode so that means uh in theory there should be two solutions there are there are two solutions there are two solutions why because this is um second order so there are two solutions so let's talk about uh the special case this first we gonna consider the case where fx is uh zero that's the easy one uh so let's
3:22:51
talk about that so that's what you call homogeneous second order equation homogeneous
3:23:00
homogeneous uh second order uh difference liquidation homogeneous second order differential equation so uh what is that that is means if if x function is identically zero then the differential equation is homogeneous it's just a homo homogeneous okay in short sometimes uh good and now we have this a very important result uh what we call the superposition principle so what is so this is what we call the superposition principle uh so let's say superposition
3:23:49
so what it says that if y1 x and y2 x if y1 x and y2 x are two solutions our two solutions are two solutions of the homogeneous equation of the homogeneous equation
3:24:19
uh if the homogeneous equation
3:24:24
y double prime plus p x y prime plus q x equals 0 then y equal c 1 y 1 plus c 2 y 2 is also is also is also a solution it's also solution for any for any c1 and c2 so that's very interesting result so what it says that if you know that y1 and y2 are solutions then for any c1 and c2 y equals c1 y1 plus c2 by 2 is also a solution okay so um so this is a very interesting result and let's prove this one so i'm going to prove this one so we are not going to prove many in a
3:25:19
review but this is like one of the very interesting result and you need to know this one so i'm going to prove this one so what we normally do is we can write the differential equation l operator so uh let's say define l that's the operator so that means the second derivative that means you need d squared dx squared plus this one we're going to say px d over dx plus qx so we can define this operator so um and then what can happen this
3:25:52
particular differential operator we define this difference operator and then we know that this differential operator is a linear operator so linear operator means if you have a sum you can write as as 2 and if there's a constant you can put it out that's what we call a differential linear operator so this is a linear operator okay so let's try this this is a linear operator this is a linear operator now what we can do we're going to act on that so uh so we know that y1 and y2 are solutions
3:26:31
that means if you operate on l certainly l y one and l y two are zero because that's uh because y and y two are solutions so what we can do we're gonna uh operate on that so uh so let's start it like this yeah i'm going to write this extra one so this operator is linear there is there are two results the first one l y one plus y two equal l y 1 plus l by 2 that means your sum you can break it like that that's the first condition of a linear operator
3:27:21
the second condition is if there's a constant and y 1 you can pull the constant out so that's the second operator so that's what we mean by a linear operator linear operator so now let's use that fact so what we can do we need to consider l y when you consider l y so then considering
3:27:48
considering l y was what is y y means c one y one plus c two y 2 now we know that l is a linear operator so you're going to break it so we can add this one l c 1 y 1 plus l c 2 y 2 using the first property this is from the first property now we're in the second property our second property second property says you can put the c out so l y one plus l y two sometimes you just write it like that without even bracket this coming from the second property but what we know now we know that since a y
3:28:30
one is a solution this has to be zero since y two is a solution this has to be zero so zero plus zero that means the whole thing is this says that l y equals 0 that means this implies that y is also a solution so this implies that so therefore therefore ah therefore y is also a solution so that's the proof so very simple proof are using linear operators uh so that's what we call the superposition principle so let's try to grow to um the linear independence linear independence
3:29:19
so what's the linear independent means let's go with the definition this is a actual definition so if c one y one function of ah y minus function x plus c two y two equals 0 0 for n x 4 n x implies that implies that this can implies that c one and c two are zero that means only the trivial solution if it implies that we call we call y one and y two are linearly independent linearly independent linearly independent this because we are only considering two solutions so we have two
3:30:12
but we can just extend this one for three solution four solutions like that i just we just write for 2 we are linearly independent if the function are not linearly independent if the functions are not legally independent if the functions shortcut form if the functions are not not linearly independent
3:30:39
the function or if the function are not linearly independent ah they are they are linearly
3:30:51
depend because you really depend so what normally happens is we start with the with general set if there's a linearly dependent one we're going to remove one one of the dependent one now i mean because the new set see whether this is linearly independent or dependent if it is dependent again we remove one linearly dependent one we keep doing it at one point what's gonna happen you get vectors with the circle vectors actually they're gonna be linearly independent so we can do that
3:31:24
like reduction method we can remove always try to remove a dependent one then i we end up with the set which is linearly um independent that's the kind of criteria so let's try to do like one example just to see how things can work so that show that
3:31:47
uh y 1 x equal x that's the first function and the second function y2x is let's say cosine x linearly independent
3:32:09
linearly independent so what we are going to begin is the definition ah so the answer so let us say suppose ah we have c 1 x plus c 2 cosine x equals 0 for any x for any x now what we can do because this should be true for any x so we can just pick any value we like so we're going to say x equals 0 so substitute
3:32:44
x equals 0 so this of x equals 0 what will happen we get c1 times 0 plus c2 times cosine 0 equals 0 to be true for any x so we put x equal 0 so this is 1 so this says that c 2 equals 0 so now we prove that c two equals zero now what we can do so ah so that is this implies that solving a math problem is a logical argument so that's why we write it like that so what we're going to so c 2 equals 0. so we only have c 1 x equals now what we're going to just put put any non-zero value so now let
3:33:32
x equal one just pick anything other than zero so this says that c one equals zero so what happened so so if you start with c one x plus e two x equals zero if you start with this so we start with that that's the definition by proving the linear independent by difference we assume that and then we this implies that c1 and c2 has to be zero so that means by the definition the two functions are linearly independent okay ah so we can see thus
3:34:07
c1 equals c2 equals 0 therefore by the definition therefore therefore by the definition therefore by the definition y1 x and y 2 x which is cosine x are linearly independent okay so that's the proof very short proof but just substitute proper values and we try to close the c one equals c two equals zero so that's how we prove like something linearly independent now uh so this can't be lengthy sometimes so we have instead what we call the wronskian test so let's talk about that
3:35:00
wrong skin test for a linear independence on skin is fully independent what it says that we're going to calculate the wrong scheme so what's the wrong skin of two functions y one y two so it defined as a two by two matrix ah two by two determinant it is y one y two y one prime y two prime that's for two by two so what it says that if this is not equal to zero for sum not all for sum all is good but we don't need that for some x naught some x value then y one and y two are linearly independent
3:35:54
are you nearly independent we prove that where this is coming from so you can look at the not uh to find out where is coming from okay so let's use this idea to prove this result just to see how it works so that so that y1 equal sine 3x and y2 equal cosine 3x are linearly independent a linearly independent thought form fully independence ah so let's do the same thing so we're going to go with the wronskian y 1 y 2 which is according to the definition y 1 y 2 y 1 prime y 2 prime let's substitute
3:36:50
this is sine 3x the derivative of sine means cosine 3 comes out by the chain rule so it's a 3 cosine 3x this is cosine cosine 3x by the chain rule it is negative 3 sine 3x and now what we're going to find the determinant is a 2 by 2 determine we multiply the main diagonal and subtract the product of that diagram so if you multiply the main diagonal you get negative 3 sine squared 3x minus you multiply the other diagonal so you get 3 cosines squared 3x
3:37:39
now what happened you can pull the negative 3 out so you get sine squared 3x plus cosine squared 3x we know that by the pythagorean identity this is 1 you simply end up with negative 3 this is automatically not equal to 0 for any ah for any x y for any x there's no x that means independent that is f you can say for nx therefore those two functions are linearly independent so therefore therefore y1x which is sine 3x and y2x which is cosine 3x are linearly independent you can see that
3:38:33
it's very simple uh just need to uh use the definition uh the the definition of the wronskian and then find the two by two determinant if you forget what's the two by two determine is if you have a b c d the determinant we use the vertical bus is a d minus b c that's the two by two a determinant okay so now we have um there are like kind of common sets uh which are linearly independent so i'm going to give you that set because it's it's good to remember that
3:39:10
not just for this class but in general so let's let's talk about that so it's a result
3:39:23
the following sets and their subsets following sets so if you pick any subject still work following sets
3:39:34
following assets and they are subset you can pick any subset and they are subsets you can just print number of uh elements that's what it means following sets and their linearly independent
3:40:02
so this is like kind of good set to remember so let's start i'm going to give like about seven of them here but there are more so the first one is this set one x x squared x cube like that this set is linearly independent and if you pick any subset for example 1 2 those are linear independent let's say you put 1 x squared x 3 those are independent that's why i say any subset you can this whole set is independent and even if you pick any subset of that
3:40:32
still linearly independent because if you make it small it's always independent problem comes when you add more ah the second one is one i just add one to all of them uh ex then you square this that means e2x squared is e3 x not the square you multiply by and x so like that so this set is line independent 1 e x e 2 x e 3 x like that ah actually you can just go with any general one so for example let's write that so i can add one more so it's a new one
3:41:09
e alpha like alpha is any nonzero number e 2 alpha e 3 alpha like that so this set is linearly independent number 4 ah how about this one 1 x this is a very useful one we're going to use this a lot e alpha x and then we're going to multiply the two so x e alpha x and then x squared e alpha x this set is also integral you can just multiply by x every time so ah so this set is linearly independent and any subset of this is also really independent
3:41:53
that means x uh know for example you can start with e alpha x x e alpha that's like kind of common one so we need this later uh when we talk about repeater roots so e alpha x and then you when you multiply by x is scaling independent meaning another one independent this is like useful very useful uh let's go number five uh
3:42:20
this one e alpha one x e alpha 2 x like that e alpha 3 x those alphas are different so alpha 1 does not equal alpha 2 and not equal to any of the other ones so actually this only says alpha 2 alpha 3 but it's also we have it's not equal to alpha 1 as well so if you unwrite it like that you have to write it like that actually so that's the idea so all the alphas are different and then we have sixth one one sine x cosine x and then we have sine 2x and then we have a cosine 2x
3:43:12
like that so this is very useful in fourier transform and that's the fourier transform basis actually and then we have our number seven we can square or multiply we have one and then we have sine x uh cosine x and then we can square them so sine squared x cosine squared x and even you can have any product of the previous ones for example you can have sine x cosine x these are all linearly independent and any subset of these independent and as the eighth one uh i'm gonna write this one x sine x
3:43:53
and then x sine x like that uh x squared sine x you get the idea okay like that so those are all these all those sets are linearly independent and it's good to remember okay then we have this major result um so let's try this as a theorem uh what we normally call the existence uh theorem ah so we can say ah
3:44:30
if y one x and y two x ah two solutions are two linear independences r two linearly independent this is very important it has to be linearly independent it's not only independent this is not true okay so if y1 x and y2 x are two linearly independent solutions total individual solutions of the uh homogeneous equation
3:45:07
of the homogeneous equation y double prime plus p x y prime plus q x y equals 0 the general solution that means any solution you can write in this form the general solution the general solution is given by is given by y equal c1 y1 plus c 2 by 2 they are c 1 and c 2 are any
3:45:54
constants okay these are any constants so for any constant c one and c two um the general solution is given in this form for that y one y two has to be linearly independent so that's my any solution you can write in this form y1 c1y1 plus c2y2 that's like really powerful result okay so that's about the major things and then ah so let's talk about the uh the non-homogeneous case so that's uh as we talked that topic we can talk about later but we're gonna put it here
3:46:33
so that it does like everything together so now this is all for homogeneous but now let's talk about non-homogeneous equation so the non-homogeneous equation
3:46:50
so what's an unhomogeneous equation we have y double prime plus p x y prime plus q x y equal f x so we are f x is not identical it can be zero for some values but not identically zero so the question is um how to solve it so that's the question how to solve so what we can do we're going to split the problem into two parts okay so let's talk about that so what we can do ah so again how to solve so what we're gonna do we're gonna split the problem in two parts we're gonna do a y double prime
3:47:38
plus p x y prime plus q x equals zero we're gonna get this part and that's what we call the complementary problem and then we're going to talk about the whole equation again y prime plus q x y equal effect the whole equation what we're going to do from the first one the homogeneous equation we're going to find the general solution so find the general solution of the homogeneous equation find a general solution we know that this general solution you can write as
3:48:20
we're going to call this the complementary solution we can call y c we know that the general solution of this one has the form c one y one plus c 2 y 2 for two solutions so this one is what we normally call the complementary source that's able to see here
3:48:41
complementary solution
3:48:46
and then from the other one we find one solution find a particular solution find a particular so this is just one here we find general solution here we find only one solution uh so this solution what we normally write as yp because yp that's the one solution the particular solution so this one is what we call the particular solution
3:49:21
and then the general solution of the non-homogeneous equation so the general solution general solution of the non homogeneous and homogeneous equation is given by y is the sum of the 2 y c plus y p which is c 1 y 1 plus c 2 y 2 equal y p and we know that how many conditions how many unknown we need how many c's we need because this is second order differential equation that means we need two and you can see that those are those two unknown uh so this is the general
3:50:08
solution with two unknown and then we can find those c one c two using the initial conditions okay so uh so that's the that's the form of the general solution of the a second order non-homogeneous equation
3:50:27
okay ah to find yp we're going to use the two methods so let's talk about write that to find by p we use two methods the first one is the undetermined coefficient method
3:50:55
that we discussed later in the discussion and then we also use the variation of parameters method
3:51:14
the problem is fused undetermined coefficient method there are only three cases that we can do there are only three cases we can talk about those three later but uh we can have the variation of parameters in that case uh the only problem is you might get an integral that you cannot do then you're going to get stuck again but otherwise you can use that again to find c1 c2 we use initial conditions okay that we normally do at the end to find c1 and c2 use initial conditions
3:51:56
so we use the initial conditions at the end so remember that it's kind of joke but that that's how you remember initial conditions you use at the ends okay so that's about that uh and then we have this uh existence theorem for uh second non-homogeneous equations so let's write that so it's a theorem uh existence and uniqueness theorem
3:52:32
so what it says that if if p q and f are continuous are continuous
3:52:47
are continuous
3:52:52
uh on some interval
3:52:59
on some interval containing a
3:53:04
containing a ah then then by double prime plus px y prime plus q x y equal f x has a unique solution that's very important has a unique
3:53:30
solution a satisfying satisfying y a equal b and y prime a equal b1 uh on that interval i so what it says that uh if that's the a so on this interval this is the interval i so on that interval it says that there's a unique solution because the the where is that interval coming from because that's where those three functions are continuous so that interval there's a unique solution so it's very interesting result okay so let's just do one last example um
3:54:14
from what we discussed so far so two solutions two solutions of x double prime minus x prime minus 12 x uh are given by at will x equals 0 equals 0 are given by are given by x 1 equal 2 e negative 3 t and x 2 equal negative 5 e 4 t so that so that they are independent so that they are linearly independent we can use the wrong scale test for that y 1 y 2 ah not y 1 y 2 is x 1 x
3:55:28
so we have x 1 and x 2 so it's x 1 x 2 x 1 prime x 2 prime so if you plug in 2 e negative 3 t we take the derivative negative 6 e negative 3 t and then we have negative 5 e 4 t so negative 2 any e for t uh let's take the determinant again multiply this first and subtract that so what do you get if you multiply that one you're gonna get negative 40 e negative three t times e to the four t and you when you multiply the other one you get positive and again minus because of uh it's a b a d minus b
3:56:26
c so you get 30 uh e to the negative 2 3 t times e to the 4 t when you multiply you get negative 7 e e to the t which is not equal to 0 for any t the only one t but so for n t doesn't equal 0. so therefore
3:56:52
therefore x 1 which is 2 e negative 3 t and x 2 which is negative 5 e 40 are linearly independent
3:57:18
good so that's the proof a very simple so ah so that's the uh about basics about the uh second order differential equations now let's move on to the reduction or the reduction of order method
3:57:39
so in this uh we're going to talk about the reduction of order method that's number 17 a reduction of order method reduction of order method so what's going to happen uh you start with a higher order differential equation and you're going to bring one level down so that means uh second order differential equation uh second order differential equation you can write as a first order differential equation that's the idea it's like a reduction of order whatever go down
3:58:19
so how it works ah one solution is given one solution is given uh find the other solution find the other solution that's the kind of idea find the other one find the other one give um so how it works so uh let ah y one which is not equal to zero be a solution okay that one we know okay we know this
3:59:09
so we know this one now what we can do we can assume a solution so we can assume assume a second solution assume a second solution ah of the form
3:59:32
of the form y equal as a second solution let's call it y 2 y 2 equal the y1 solution times u so u is unknown that's the one we need to find
3:59:51
uh the unknown function we're going to find that unknown function so the goal is to find this okay that's the starting point we're going to assume a solution of the form y1 times u goal is to find you okay goal is to find it so the goal is to find u because after you find u then we know the second solution okay but how we do it we can start taking derivatives and then plug into the equation and then solve so that's how we do it
4:00:31
so uh so we start with a taking derivative so we have this one and then we can take the so take find u so what we're going to uh take derivatives
4:00:48
take derivatives and substitute
4:00:57
uh into our formula so what is that so uh so difference equation is y double prime uh plus p x y prime plus q x so we're gonna uh q x y
4:01:14
q x of y equals zero so we're gonna plug in here so we can take the radius and plug in here so let's see how it works now um
4:01:31
so we have we have y2 equal y1u so take derivatives so that means y2 prime by the product rule uh this can be y one prime u plus uh u prime y one that's what we get from the product rule now uh let's take the uh derivative again so if you take the second derivative by two double prime now we have two so you get four terms so we can take the derivative so this is y one double prime u plus y1 prime u prime from the other one u double prime y1 plus u prime
4:02:25
y1 now you can see that you can combine these two and write one equation so you can write this one as y1 double prime u plus 2 uh u prime y1 prime and then we have u double prime y1 good now what we're going to do we're going to substitute this okay because we know that this is solution since uh y2 is a solution of y prime p x y double prime plus p x y prime plus q x equals 0 we can substitute here so we're going to get 0 equal uh we have i 2 double prime plus p x y 2 prime plus plus
4:03:31
q x y 2 so let's plug in for each term so what is y 2 double prime that is y 1 double prime u plus 2 u prime y1 prime plus u double prime y1 that's the first one it's coming from here now px times px times u2 prime u2 prime means u1 uh y1 prime u plus u prime y1 it's coming from here
4:04:15
now the other one which is qx times ah y2 which is y1u but we know this whole thing equals 0. now what we're going to do we're going to kind of collect the terms in a different way so that we can use the fact that y1 equals 0 y 1 is the solution so what we're trying to do we're going to isolate the u terms you know we're going to go with the u double prime terms first so what are u double prime terms uh u double prime we have y 1 uh u double prime
4:04:51
that means this term then we're going to go with the y prime terms whether by prime terms uh let's collect all the y prime terms we have 2 y 1 prime plus that's that term and then we ask the other y prime so this is another y prime term so let's collect those two together so we have p x um y1 so those two have a y prime term plus um u prime terms and then let's call the u terms well u terms uh there's a u term here that's your term here this other u
4:05:35
term let's combine all of them and take you out as a factor so you're gonna get y one double prime plus from the second one p x y 1 prime plus q x y 1 that's what you get from the u terms and then take u out but we know that y 1 is a solution that's we know that means this has to be 0 because y 1 is a solution so that means that i'm going to go away so you simply end up with this expression y1 u double prime plus 2 times y1 prime plus p y1
4:06:26
u prime that's what you have so that means that has to be equal to 0 because that's equal to the sub so that's the that's the whole argument that's what we're going to do you can assume a solution take derivatives and plug in and you can see that there's a special term that's going to go away because y1 is a solution and then you simply end up with the new expression so what we have now so this is the interesting part so we only have this one so we have y1 u double prime plus 2 y 1 prime plus p
4:07:01
x y 1 times u prime equals 0 so that is the equation we end up with now the goal is to solve this for you solve for u but you can see that there you you prime is there not u u prime so what we can do we gonna make a reduction of order here so what we're going to this is the whole point so we're going to say let v equal u prime if you say equal u prime since there's no u term here uh then we have the prime equal u prime u double prime
4:07:45
so what happened then we can rewrite the above equation in this form so we can write the the equation like this we can say uh y1 v prime plus 2 y1 prime px y1 u prime means v equals 0 now you can see that this is a first order linear equation first order linear equation so first we can do we can divide by this term if you divide by that term because that we know that that not equals zero so we prime plus if you divide by that you get uh what do you get
4:08:27
you get 2 y 1 prime over y 1 plus p x just p x e times v equals zero so this is first order linear in v this is first order linear in v now we can solve this we can use the integrating factor method or some other techniques to solve this one because this can even be uh because you can factor this you can this can be separable so depending on what is y1 looks like so that means uh so the next step is our solve for v first
4:09:12
and then and then solve for u because after you solve for v then we can use the fact that v equal u prime you integrate one more then you're gonna get solved for you so that's the idea okay so let's do the actual calculation uh in a problem because what happened when you do a real problem there are a lot of simplification going on so it won't be this lengthy when you do the actual problem so let's do a problem so and also i'm going to give you the formula for that so you can
4:09:50
uh use that formula in the exam but this is just a method if you want to do something late ok good so let's do this one so it is given that it is given that it is given that y1 equal x squared is a solution of a solution of x squared y double prime minus 3 x y prime plus four y equals zero so we don't know how to solve this equation but somehow you can use a trial and error argument or something we know that this is a solution the goal
4:10:38
is to what are the other solutions find the resolution okay so find the second solution because there should be two solutions because the second order find the second solution so that's the goal okay so we can just reduction forward argument okay because we know that that solution so let's see the argument and completely solve this problem um good so the answer assume a solution of the form assume a solution of the form
4:11:30
y2 equals x squared you because we know we so we don't write y 1 x square u now what we do we take derivatives okay so taking derivatives
4:11:47
we have y prime equal is the product rule so it is 2 x u plus x squared u prime we have second derivative in the equation so that means in the second derivative as well here so if we do the second derivative you get 2 u plus 2 x u prime from the first term from the second term you get 2x u prime plus x squared u double prime so you can combine this and you can write this on as 2u plus 4 x u prime plus x squared u double prime we combine these two
4:12:38
okay now uh since uh so actually y two essence by 2 is a solution so we can substitute that to the equation which is write a solution you can substitute the equation so we're going to get x squared y 2 double prime minus 3 x y 2 prime plus 4 y 2 equal x squared times substitute y 2 prime it is 2 u plus 4 x u prime plus x squared u double prime this is simply coming from here minus 3x y prime y prime means 3 x u plus x squared u prime plus 4 um y 2 y 2 means x squared u which is equal to
4:13:52
because that's the assumption because i think there's a solution now let's see how it's going to simplify you can see there are a lot of simplification uh going on here there's a typo so this is not three this is two uh you can see that there are a lot of calculations cancellations going on uh let's look at the x squared u term so there's x square u term here this x squared u term and then we also x square u terminal and you can see that those three get
4:14:25
cancelled so those three get cancelled out uh because we have two four and negative six it's gone so that means you only end up with uh just a few other terms so you have uh
4:14:47
so you're gonna get uh it's what you're gonna get you're going to get uh there's x cubed term here so this is x cubed term because that's with x x cubed term there's another x cubed term here so you're going to combine those 2 x cubed term and there's x 4 so this is x fourth term because there's x squared so that means you have x cubed and x fourth so this can uh give you that uh x fourth u double prime plus x cube u prime equals 0 so that's what you get everything else
4:15:26
get cancelled now what we can do is um we can divide this by x fourth so if you divide by x fourth so we can divide by that term uh so that is gonna be u this can be u uh double prime plus one over x u prime equals zero now what we to do we're going to use a substitution so we're going to say let v equal u prime so let uh v equal u prime then we have um v prime equal u double prime so we can plug in so our equation simply becomes v prime plus 1 of x v equals 0 so you can solve this problem using
4:16:09
separable method or the integrating factors method so you can use any measure so it depends on the problem uh sometimes you can do that something you can't do but most of the time uh yeah so it's always like in this problem you can use both of the techniques so we can use the integrating factors for this one it's kind of straightforward so let's do that and finish the problem so we have our integrating factor i which is the integral of p x dx so it's integral of p x p x is 1 over x
4:16:45
dx so we have e ln x and then e l and x cancel out so you get x so that's what you get now ah from the theory from the integ uh theory of integrating factors we know that this side is simply the derivative so you're going to get d over dx the v times integrating factor equal on the other side is 0 so integrating
4:17:23
so integrating
4:17:28
integrating we have vx equal a constant i'm going to call c1 so this says that we equal c1 over x and now we're going to substitute for v but v means u prime so this is that u prime equal c one of x to get u we can integrate so u equal the integral of u prime calc 1 as integral of c 1 out or no x dx that means c 1 l n x plus c 2 but what you can do ah you can ignore c2 because we are looking for one solution we can even say c one equal one so set c one equal one
4:18:16
and c two equals zero because we are looking just one solution here so that means you're going to get u equal ln x uh so x actually here you can say absolute value because uh we did not restrict only to positive x so x can be positive and negative so you can have linux absolutely so okay so that means this is second solution so then the second solution y2 equal x squared u uh which is x squared ln x so that's how you find the second
4:18:55
solution using the reduction of order method so it's not that difficult but what we did was we start we assume a solution of the form y1 u substitute that there's a nice simplification you can see that there's someone that there are a lot of simplifications a lot of terms get cancelled out and then you get a simple equation that you can solve using the separation of variable or integrating factors method so now i'm going to give you the
4:19:20
reduction of order formula so you can directly use this formula if you get a problem like that so the theorem ah the reduction of order method reduction
4:19:35
of order formula so let's say if y1 x y1 x not t equal to 0 is a solution it's a solution of y double prime p x y prime plus q qx y equals zero remember it has to be of this form we not of this form you have to bring it to this form okay we need that we need to isolate the p x term so sometimes you have something in front so i have to divide by that first okay so it's very important it's a solution of this then
4:20:20
another solution in another linearly independent so this automatically independent another linearly independent
4:20:34
another linearly independent solution is given by is given by y 2 x equal y 1 x integral of e to the negative e x dx let's say you need to know what the p x and divided by y one x whole thing square sigma square by one dx so that is the formula that's a reduction of order formula and you can directly use that so let's use that in a problem and see how the whole calculation works so this is a problem of that type so if it is given that if it is given that y1 equal x is a solution each solution of
4:21:40
x squared y double prime minus x y prime plus y equals zero find
4:21:55
the generous solution find the general solution so ah so asking for general sources that means this is the second node the difference is equation there's another solution so we're going to find the other solution using the integrating factor method so the problem is when you look at this one this is not of the standard form that's x squared term so we'll divide by that so we have ah x squared y double prime minus x y prime plus y equals zero so what we can do
4:22:26
we can divide by x square to bring it to standard form okay so the standard form
4:22:37
standard form is if you divide by that you're going to get y double prime minus 1 of x y prime plus 1 over x squared y r equal 0 so you can see where p x is negative 1 x that's the p x okay so you have to isolate the p x term uh good so uh then we can use the formula to find the second solution the second solution is
4:23:22
the second solution is given by by the formula it is y2 equal y1 times e to the minus px dx over y 1 squared d x so let's plug in here we have everything we know so y 1 is x the integral of a to the minus p x is a negative one no x d x over x squared d x this you can simplify a little bit so you get e to the one of x dx over x squared and you can see that you cannot bring the x square out because as you're still integrating inside uh
4:24:19
so from that you can see this is x the top is simply e to the ln x of x squared dx and then we know that when you have e l and x uh e l and x e l n uh cancel out so you need x so this is x times x over x squared dx this is simply 1 no x dx 1 x means ln x so we're going to say x ln x and we are not going to have to see here ah the reason is we are looking for one solution okay so that's that ah now the next step is we need to show that these two are linearly dependent uh
4:25:14
okay to show that
4:25:20
ah they are linearly independent actually if you remember the theory it says that these are say another linear dependency so either you say that independent solution by the theorem or you can just prove this this is not that difficult yeah so the solution is automatically independent the solution that we get we can prove that taking a you're using this formula but it's also not that difficult to show that so let's plug in here and see what happened this x derivative of x is 1
4:25:53
we have x ln x uh derivative that is use the product rule so you get ln x plus x times 1 of x even in this case uh so let's simplify when you simplify what happened you're going to multiply uh so what you get here you get x 1 x ln x uh ln x plus 1 that's what you get so we can multiply through when you multiply through what you get you get x ln x plus x minus x ln that's a ticket you can see these two get cancelled so you get x which is not equal to zero
4:26:43
in general where x can be anything other than 0 so that means yes not equal 0 so they are linearly independent so
4:26:58
y 1 equal x and y2 equal x ln x are linearly independent good ah therefore the general solution is because in the theory we know that those two has to be linearly dependent to find the general solution otherwise it's not going to be so therefore the general solution is is given by
4:27:40
a y equal c 1 y 1 plus c 2 y 2 equals c 1 x plus c 2 x ln x so that's the general solution so that's how the uh reduction of order method uh works so it's uh very interesting and not that bad okay so if you have a problem like that you can directly use the reduction over the formula okay let's move on to the most important uh topic so what's the most important topic the constant coefficient differential equations so number 18 is the most important one the constant coefficient
4:28:32
constant coefficient uh second order uh constant coefficient second order differential equation
4:28:46
good
4:28:50
ah so this kind of equation has this form so as the form
4:28:58
has the form a constant times y double prime another constant time y prime another concern time y equals zero equal to the homogeneous equation and then we know that a b and c are constants that's the case we trying to um understand so that is the case very special case good now what we're going to do how to solve this one uh what we did we assume a solution of the form y equal e to the lambda so that's the that's how the solution work so assume a solution of the form
4:29:47
of the form y equal e to the lambda and then what we can do you can take the derivatives of that and plug it we assume our solution of the form and then we can take the derivatives okay so take derivatives so we're going to get y prime which is lambda is a constant lambda e lambda and then we can take another derivative double prime so we can take lambda squared e to the lambda okay so we know how to take this like that oh actually it's not e to the uh it's lambda t it's not just lambda
4:30:23
so uh the e to the lambda is a constant it's not a function it's just one is a constant for e to the e lambda t uh and then also here uh so t and t okay good um yeah so we assume the solution of the form and take the derivatives so y equal e to the lambda t now we're going to substitute this in the equation so let's do that uh substitute so if you substitute you're going to get a lambda squared a to the lambda t that's the first one plus the second term b
4:31:09
a lambda a to the lambda t that's for y prime because this is y double prime this is y prime a plus c e lambda t equals 0 this is y so we're gonna substitute 4 3 now you can see that e lambda t is common so you can take it out as a factor so if you take as a factor you're going to get a lambda squared plus b lambda plus c equals 0
4:31:47
for any t but we know that e to the lambda t never becomes 0 since e lambda t not equal to 0 the only option that this can be 0 is a lambda squared plus b lambda plus c equals 0 this is what we call the characteristic equation of the differential equation because this has to be satisfied if e lambda t is a solution okay so this is what we call the the characteristic equation
4:32:29
characteristic characteristic equation good and you can directly write this equation so what happened when there's a derivative it's a second derivative you get lambda squared it's the first derivative you get lambda 1 there's no derivative you can lambda 0 that means constant so you can directly write you don't want to show this work here you can directly write the characteristic equation if you have problem like that okay then let's try this important theorem uh so what it says that
4:33:01
okay so so the theorem
4:33:09
so consider
4:33:13
the differential equation uh a y double prime plus b y prime plus c y equals zero we are a b c constant
4:33:34
with
4:33:38
okay so yeah and then ah then the characteristic equation then the characteristic equation we call cha equation characteristic equation is simply a lambda squared plus b lambda plus c equals zero so you seem to replace the derivative terms by lambda powers so see what happens is there are connection y double prime lambda square y one prime lambda one power there are no derivatives or no lambda powers is lambda zero powers okay now we can write the general solution for this one
4:34:16
in these three forms so let's write them separately so the general our solution has the form ah and let's see and the generous solution and the general solution has the form let's try it one by one ah so there are three cases to consider because this is a quadratic equation uh so it can be uh so there can be two real roots distinct real roots equal real roots or complex roots so we have three cases uh so if b squared minus 4 a c positive
4:35:07
then what happened the lambda 1 and lambda 2 are real distinct
4:35:20
roots so general solution has this form y equal c 1 e lambda 1 x plus c 2 e lambda 2 x in that case the second case the second case if b squared minus 4 ac equals 0 then what happened lambda 1 equals lambda 2 ah then what happened this is a real repeated situation real and repeat it
4:35:56
real repeated so the solution has to form c1 plus c2 x times one of those terms ah lambda 1 x oh you can expand it and write it like that c 1 e lambda 1 x plus c 2 x e lambda 1 x you can write it like that oh it doesn't matter which form okay and the third case if the b squared minus 4 is c negative now what happened your lambda can be written as alpha plus or minus a beta i complex conjugate this is what we call the complex conjugate situation
4:36:56
complex conjugate so in that case what will happen your y has to form c 1 e alpha x cosine beta x plus c 2 e alpha x sine beta x so that's what happened in the in that case uh we have a cosine and a sine term so those are the uh three cases um that we want to remember so again what happened in the first case you get real distinct real repeated and then complex conjugate and you can directly use the result so let's see this in a problem or two so let's look at this problem uh find
4:37:48
the general solution of find the general solution of uh let's say a 2 y double prime minus y prime plus 3 y equals 0 so if you can provide that the first thing is we look at the characteristic equation so the characteristic equation
4:38:24
equation is simply 2 lambda squared minus lambda plus 3 equals 0. this is very easy to find factors so what are the factors uh 2 times 6 so you have 6 you're going to bring write this as a product with the add the 2 you get negative 1 so that means it has to be negative 3 and 2 so that means your factors are 2 lambda and lambda now to find the number so you know that 2 so how to get this number you simply divide that 2 by this 2 so that means it is
4:39:11
plus 1 and for the other number you divide negative 3 with 1 so that means negative 3 so those are the two numbers you get and if you don't know how to do that then you can look at the factoring polynomial video and i explain all different type of polynomials in there quadratic cubic quintiquatic all of them so that's a video that you can watch okay good uh so then we have the two roots so then you can see so then uh you're gonna set them equal 0. so you're going to get
4:39:50
lambda equal negative 1 and lambda equal 3 halves you set them equal to 0 and get the two values if you want i can write that step as well so what we normally do is after that so you're going to set them equal so then we have 2 lambda minus 3 equals 0 o lambda plus 1 equals 0. so this says that if you bring this to the other side 3 and divide by two this says that lambda equal three halves or lambda equal negative one so oh it's not an end we don't know whether we need both good
4:40:43
so those are two solutions of lambda so the general solution is general solution is given by ah y equal according theory it is c 1 e lambda 1 t plus c 2 e lambda 2 t so let's plug in the two values so this e c uh c1 i normally put the smaller one first it's a negative one so negative t plus c two e the larger one three halves t so that is the general solution of the differential equation very simple so let's look at one from the other type
4:41:36
and here i pick the independent variable as x as t so it's just up to you be careful uh when you use that here you can pick x or t because i did not say anything okay good let's do uh the next example so problem number b uh so we have four x double prime minus four x prime plus x equals zero and you can definitely cannot pick x as the independent variable now because that is the dependent variable okay so the uh characteristic equation is 4 lambda squared minus 4 lambda plus 1 equals 0
4:42:26
uh how to factor this and you can see that this is 2 lambda squared minus 2 times 2 lambda plus 1 squared so this is a complete factorization problem because we know that we have formula if you are a squared minus 2 a b plus b squared this is simply a minus b whole thing squared that's exactly what we have here uh so that means this is simply 2 lambda minus 1 whole thing squared yeah you can see this again in the uh factor polynomial video good so we have that now what we can do
4:43:05
we can set this equal to zero uh so we have two lambda minus one equals zero this says that lambda equal one half why bring one to other side divided by two and then also is repeats uh because or you can see with multiplicity 2 with multiplicity
4:43:30
a 2 because it's a repeat you can see square good now uh so we're going to get the solution uh so the general solution so then the general solution
4:43:45
then the solution is y equals c1 plus c2 uh here variable your independent variable has to be t because you can pick x there so t times e to the lambda 1 t because it's a repetition ah so that means c 1 plus c 2 e c 2 t ah e to the ah one half t so that's the solution c one c two constants so let's do a last one from the other type the complex type so let's say y double prime minus y prime plus y equals 0 again the characteristic equation is lambda squared minus lambda plus 1
4:44:42
equals zero you can see that you can find factors of this one so you have to use the uh formula uh quadratic formula so this is lambda a negative b plus or minus b squared minus 4 ac square root over 2 a so let's plug in here so you get 1 plus so minus 1 minus 4 a c so the circuit over 2 a means 2 so which is 1 plus 4 minus negative 3 square root over t you need complex numbers for that so that's going to square root 3 i so this is that and so we have lambda equal 1 plus or minus square root 3
4:45:35
i over 2 if you separate this you can write this one as one half plus or minus square root 3 i over square root 3 over 2 i so you can see that that is the alpha value and that is the beta value so the general solution is so the general solution
4:46:00
general solution is given by a y let's say x is the independent variable e alpha x times c 1 cos beta x plus c 2 sine b x so that is a format so we're going to plug in here so it is e one half x times c one cos square root three over two x plus c 2 sine square root 3 over 2 x so that is the general solution with c 1 c 2 unknown and now what we can do is we can completely solve a problem with initial conditions as well and to see how it works as the last problem of this type so let's do that uh
4:47:02
so solve the initial value problem x double prime plus x prime plus three x equals 0 with x 0 equals 0 and x prime 0 equal 22 and later we're going to solve this exact same problem using the laplace transform method and you can see how it works okay but right now we're going to use the constant coefficient difference equation technique because that's much easier so let's do that okay so the answer we can look at the characteristic equation so which is lambda squared plus lambda
4:47:53
plus 3 equals 0 and then you can see i it's impossible to find the factors so we're going to go with lambda equal formula negative b plus or minus b squared minus 4 a is 1 so c over 2 so we can simplify this is negative 1 plus or minus negative 11. so negative 11 square root over 2 if you split this you can write this as negative 1 half
4:48:33
plus or minus square root 11 over 2 i so uh that's the alpha value this is the beta value so we have alpha value of the beta value so the general solution so the general solution you can write as x t uh which is c 1 e to the alpha value t so let's say it's a function of t independent variable is t and then we have cosine beta which is square root 11 over 2 t plus c 2 e to the negative half t sine square root 11 over 2 t so that is the general solution you can generate solution 3 lines now
4:49:28
only thing what we have to do is to find c 1 c 2 using the initial conditions so let's do that so we can use initial conditions so using the initial condition uh first x 0 equal 0 we have 0 equal x to the 0 let me plug in 0 wherever you see t so c 1 e to the 0 cosine 0 plus c 2 e to the 0 sine 0 so this is that this is equal to 0 this is equals to 1 e this is we know that this is one sorry this is 1 and this also one so that means you simply end up with what you get uh c1 so this is that
4:50:29
c1 equals zero so that means your solution kind of simplified because you don't need the c1 term so that means you have so does we can write x t like this it is simply c to e to the negative one half t sine square root 13 over 2 t so before you take the derivative it's a good idea to do this because then you have only one term otherwise like it to be difficult so ah
4:51:09
now what we can do is uh we can take the derivatives so taking derivatives taking a derivative so we can use the product rule here so we have x prime t equal so take the derivative of the first term so you get one half negative one half c two e to the negative one half t no change to this term plus now it will be the other term so that means side so by the chain rule square root 2 square 11 over 2 comes out you have c 2 e to the negative 1 f t sine become cosine square root 11 over 2
4:51:58
t that's what you get now we'll find the condition so x prime 0 equal 22 gives 22 negative one half c two e to the zero sine zero plus square root eleven over two c 2 e to the 0 cosine 0 that means as before this is gone this is 1 so you're going to get what you get you're going to get square root 11 over 2 c2 so that's going to give you the c2 value so this says that see 2 equal 44 over square root 11 in other words 4 square root 11 okay that's c2 value
4:53:05
so therefore we have the solution now we have x t which is 4 square root 11 e to the negative 1 half t sine square root 11 over 2 t um yeah so that's the solution so let's talk about something more about this one you can see that in this problem this is a periodic function we have sine term that means the periodic function but you can see the amplitude that's the amplitude that's what we call envelope that's going to decrease because there's an e negative term so that's going to
4:53:46
decrease so that means if you look at the graph of this one this is very interesting so ah so if you graph this function this is how it looks like so we have what we call this any loop curve e to the negative one half t that's envelope so that's gonna decrease so it's gonna go like that
4:54:08
and then it's a sine co sine that means the plus minus so what we normally do is we're gonna reflect it like that so what's going to happen now we're going to graph the sine function inside because when you have t equals 0 this is 0 so that means you can start 0 at 0 and just the sine graph between these two envelope that's exactly what happened in this consideration so this is simply this curve is simply a 4 square root 11 e to the negative one half t that's that curve
4:54:47
this exactly opposite of that this is the negative four square root 11 e to the negative one half t um so these kind of situations are like kind of common like you know if you i have like a bob hang like that and put like a sudden force like impulse then it's gonna go in like a motion like this uh it's gonna go to this corner and to that corner so what's gonna happen because of the air resistance the uh this gonna slow down so that means
4:55:23
if you look at the graph of this one if you graph it is exactly what you get start here it's going to give a certain jump because the impulse and then it's going to slow down i'm going to go and stop like that so that's the situation so this is uh the solution you you get using uh differential equations that's a kind of nice idea okay so that's about the constant coefficient of differential equations and how to solve it there are three cases and we talk about those three
4:55:49
cases uh in the next we're going to talk about another special case what we call the koji euler differential equation
4:56:01
so in this let's learn how to how to solve the kochi euler differential equation so this is number 19. uh kochi euler
4:56:14
kochi euler a differential equation some people call it euler same things ce equation uh so a quarter equation has this form so it is a x squared y double prime plus b x y prime plus c y equals zero so that's a format it's also second order homogeneous differential equation and also linear equation because b x is the independent variable so x is the independent variable so that means this is still linear so how to solve it we can assume
4:56:56
assume a solution of the form assume a solution ah of the form assuming solution of the form uh y equal x to the lambda power and then what you're trying to do we're gonna uh sub take the derivative and plug in there and solve for lambda that's what we do and sulfur sulfur lamina so that's the kind of idea so uh if so once you do that what's going to happen uh because we're looking for two solutions so let's say if they are if they are lambda 1 which is not t equal to lambda 2
4:57:44
uh then the solution then the solution is simply y equal c 1 x lambda 1 plus c 2 x lambda 2. so that's the kind of idea so let's look at a problem and see how it works uh so example solve solve x double prime x squared y double prime minus 3 x y prime minus 5 y equals 0 uh for x positive so let's say x positive so we how it works so we're going to assume the solution so assume
4:58:42
a solution assume a solution of the form
4:58:52
y equal x lambda then what happened we take the derivatives of y prime which is uh if you take the derivatives lambda x lambda minus 1 just by the power rule and then y double prime we take one more so it is lambda lambda minus 1 x to the lambda minus 2 now what we're going to do we're going to substitute that okay so we substitute that a substitute so once you substitute what happens you have x squared and then y prime y double prime which is lambda lambda minus 1 x lambda minus 2
4:59:41
minus 3 x lambda prime 2 y prime term which is x and a lambda x lambda minus 1 and then minus 5 y 2 so those are the three terms so again so this is the y double prime term this is y prime term and this is the y term so these are three terms and you can see that uh you can simplify this you get x to the lambda in each term so we can take it out so you're going to get a lambda times lambda minus 1 minus 3 lambda minus 5 x to the lambda equals 0 but we know that x to the lambda is never zero
5:00:35
because x not equal to zero so that means only thing should happen is um the whatever inside the bracket has to be zero but you can see if you simplify that you're gonna get lambda squared you get minus lambda minus of minus 4 lambda minus 5 equal to 0 that's what we get so this is very easy to solve so we're going to look at the factors uh so what are the factors uh you can see lambda minus 5 and lambda plus 1 are the factors so this is going to give you lambda equal minus 1
5:01:14
and 5. so the general solution so general solution is y equal c 1 x to the lambda 1 x lambda 1 not x lambda 1 plus c 2 x with the lambda 2 so we're going to plug in that so this e 1 x to the minus 1 plus c c2 x to the phi if you rewrite it um we bringing x down so you can write this one as c one over x plus c two x with the phi this is 4x plus it good so that's the solution so it's it's not that difficult uh so you solve for we assume our
5:02:05
solution of x to the lambda take the reason plug in solve for lambda so let's discuss another method what we're trying to do here is we're going to transform the problem to another domain another variable so another method
5:02:22
another method so we can transform the problem so let's say for x positive what we're gonna do we're gonna let x equal e to the t so we can transform the problem to the t variable uh now what's going to happen you can show that so i'm going to skip that the how it works you can look at the video general video for that uh so then what happen once you do that you can transform the variable x to t variable okay so what happened now in variable t
5:03:03
in variable t the differential equation becomes a y double prime t plus b minus a y prime again the variable t plus c y t equals 0. you can see that this is a constant coefficient differential equation and we know how to solve this uh so that's the idea so we transform the euler coach equation to a constant coefficient differential equation by this special transformation uh so this is for x positive if we x negative we can say x equal negative e to the t the
5:03:43
interesting thing is you get exact same equation even in that case very interesting so now this is constant coefficient so this is a constant coefficient this is a constant coefficient our equation and we know how to solve that so we can look at the characteristic equation a lambda squared plus b minus a lambda plus c equals here that's what he gets so that's the characteristic equation for the coach euler under the new transformation now if you know these values then we know the uh
5:04:27
the solutions because in in the t variable what you want to do is you can transform the variable back so ah so this is the characteristic equation and you can see this is quadratic in lambda so so this is quadratic in lambda again this is quadratic in lambda that means uh there are two solutions okay so let's talk about the uh separate cases and also you can see how the how the transformation works so we can talk about that so let's go one by one if
5:05:20
b minus a squared minus 4ac that's the discriminant for this if it is positive there are two solutions uh
5:05:36
so this have two solutions so if you call them r1 r2 r1 and r2 those are two numbers you get so this is real distinct situation so your solution y equal ah in the t variable it is c 1 e r one t plus c two e r two t now what we're gonna do we're gonna transform this to uh x but we know this transformation in the equivalent transformation uh so what we have we have x equal e to the t if and only if t is simply ln x so that's the transformation that we can
5:06:21
use because the exponential you can write as a log like that so t is um ln x so let's use that that means e to the t is simply x so if you replace that then this simply you can write this on as c one e to the t is x so it's become x to the r 1 same thing if the t is x so this becomes c 2 x with the r 2 so that's the idea and why this is algebra because if your e r 1 t you can write as e to the t r 1 that means x r 1 so that that's what happened just basic algebra
5:07:02
so that's how you get the solution in x okay so that you you solve the new characteristic equation this is the characteristic equation for the uh cultural equation and then after you get that those those values whatever the values simply gonna go uh here r1 and as r and r2 so very simple but you have a new characteristic equation that's the only thing in ram that's the first case and the second case if b minus a squared minus 4 ac equals 0. what's going to happen you're going to get 2 equal
5:07:45
so you're going to get r1 equal r2 a real uh repeated
5:07:55
in this case we know how the solution ah in the t variable it has to be ah c one plus c two t e to the r one t now again the same argument c one plus c two what is t t equal l n x so now put the ln x here and then e this is again as before x to the this simply become x with the r one as before that's the second case and the third case if b minus a squared minus 4ac negative then we have as before r is complex so alpha plus or minus i beta then we can write so it's a complex
5:08:53
conjugate situation so we're going to have the solution in the new t variable so the t variable solution looks like e alpha t c 1 cosine beta t plus c 2 sine beta t but what you want to do is you can write this one using the new variable so with the new variable you're going to get e t is x as x to the alpha as before c 1 cosine no change beta t is ln x x is positive so you don't need the absolute sign here plus the same thing c 2 sine beta ln x so that's the uh those are three
5:09:50
solutions so what you want to do once you get the kojol equation uh look at the new characteristic equation find the values the lambda values lambda 1 lambda 2 we just call it r because just to remember the difference and after you get that uh you simply get the three solutions okay so that's the first solution just put the new values there new values there and the new values in the last equation which is little long good so those are three cases so this is for ex
5:10:24
expositive but we can show that so i'm going to write that just for your knowledge and for the future or for the exam ah we can say in okay um so we can see it can be shown that let's try it it can be shown that it can be shown that it can be shown that for x equal negative e to the t that's the other case that means x is negative we get the exact same character equation we get the same we get the same characteristic equation that's very interesting we get the same
5:11:17
we get the same characteristic equation uh so the general solution so the general solution for x not t equal to 0 so what you want to do we want to consider the positive case in a negative case so instead of x we put absolute value so that means the three cases are if p minus a squared minus 4 a c positive then the solution is y c 1 not just x the absolute x because it can be positive or negative r1 because we don't know whether those are integers so be careful
5:12:05
and then c2 x are two so it has the both positive and negative case only thing is you include the absolute sign and the next one is if b minus a squared minus 4 a c exactly equal to 0 in that case y equal c 1 plus c2 ln absolute value of x times absolute value of x r1 that's the common r1 solution the third case is if b minus a minus 4 a c negative in that case uh what happened we have alpha beta so that means your solution is simply uh i don't think i have enough space here
5:13:06
yeah that's right uh so it is x to the alpha times c1 cosine beta ln x plus c2 sine beta ln x so that's the general case for any x other than zero so let's just solve one equation one problem to see how it works uh so say solve x double prime x x squared y double prime minus 2 x y prime minus 4 y equals 0 for positive x so the answer the characteristic equation is
5:14:06
ah so in this case a equal 1 so we can see i'm going to write a equal 1 b equal negative 2 c equal negative 4 so the characteristic equation is simply lambda squared plus minus b minus 1 lambda minus 4 equals zero and so this is that lambda squared minus 3 lambda minus 4 equals 0 so the uh you get negative 4 and 1. so this says that lambda equals negative one and four ah so real a distinct real distinct so that means the general solution general solution y equals c 1 x
5:15:13
negative 1 plus c 2 x 4 or if you divide c 1 over x plus c 2 x to the 4th power so that is the complete solution so you can see it's not that difficult but you have a new characteristic equation so it is not uh lambda squared a lambda square plus b lambda plus c it is a lambda squared b minus a b minus a lambda uh plus c okay good so that's the uh that's about the characteristics uh kochie euler equations so let's move on to the higher order differential equations we just talked about the basics
5:15:59
of that and then we're going to move on to the non-homogeneous equation and it under undetermined coefficient method so let's talk about a little bit about the uh higher order differential equation so this is topic 20 higher order
5:16:17
higher order constant coefficient constant coefficient differential equations it's very important we do the constant coefficient only in this one ah so those equations have this form a y n power b y not the power derivative n minus first derivative like that we have a first derivative and a constant another constant of d y term is equal to zero so homogeneous so we can look at the homogeneous case good same idea you can assume a solution of the form assume a solution similar solution of the form
5:17:10
y equal e lambda t then we get the characteristic equation as before so the characteristic
5:17:28
the characteristic equation is a lambda n power b lambda n minus first power like that so you could see lambda plus d equals zero so that's the characteristic equation now what we're going to factor is get the numbers and plug so uh we're gonna factor this okay so you're gonna factor this so there are a few cases so let's go case by case so we have cases the first one is if all lambdas are real and distinct that's the most common case so what's happened just the extension of what we
5:18:27
learned so the c one e lambda one if you use the x as the independent variable like that c2 e lambda 2x like that c n e lambda n x so that's what happened in the uh case so let's look at the uh repeated case very similar so the case 2 let's say if lambda 1 are repeats m times lambda 1 ah repeats m times and the rest are distinct the others are distinct
5:19:22
uh if that is the case uh then what happened why so this is extension of what we have because when we have 2 we have c 1 plus c 2 x so if you are m you're going to go m minus 1 okay that's idea so that means c 1 c 2 x like that and continue that and then we have c m m term has the power m minus 1 and then we have this is similar so l lambda 1 x plus the rest we know i just put some numbers so let's say d e to the next lambda value just
5:20:05
repeat and then the last lambda value lambda in x like that okay so the interesting part is here how we're going to write that term you can see if it is a m repeats you're going to see m minus 1 which is 2 you can see x it's like that okay and then the last case is the complex case so let's say if uh coefficients are real so it's very important coefficients are real if the questions are real then ah
5:20:48
then we have complex conjugate roots then we have complex conjugates always complex conjugate roots if the if the covalent are real the solution has to be complex conjugate that's like theory we proven algebra uh complex conjugate
5:21:15
let's say complex conjugate complex conjugate pairs
5:21:24
so let's see uh lambda equal alpha plus so minus i beta the whole thing we will go with pairs okay repeats repeats m times this thing can repeat m times and others are distinct others are distinct so in that case what's going to happen y is simply c1 plus same thing c2x like that so we have c m x m minus 1 that's the coefficient of e alpha x cosine beta x plus a similar term for cos as sine you have d1 plus d2 x like that the m x m minus 1 e alpha x
5:22:35
sine beta x and then plus the other terms all the non-equals but see the format so we have we have 2 there so that means we already have two m off okay because there are two m mini terms and then the rest rest is n minus two and many terms are there and that's gonna follow the other pattern okay so that's the idea and then uh let's talk about the linear independence linear uh independence
5:23:11
is very similar so we can write the wranskian
5:23:18
uh so for example if we have three functions so we can extend that to many but let's just for three so we have the wranskian so it's going to be y 1 y 2 y 3 take the first derivative
5:23:36
ah so this is 3 take the first derivative and take the second derivative
5:23:48
so that's a wranskian and then if this is not t equal to 0 in general for some x naught the functions are then this implies that y1 y2 and y3 are linearly independent okay so that's the kind of idea so very similar ah so as an exercise you can try this um you can see the detail in the actual lecture so for example show that so that 1 sine x cosine x these three functions are linearly independent
5:24:42
and we already know that this is a one of the uh common linear independent set we discussed that earlier so just prove it and you can see that this is very easy because one is there once you take the first derivative is zero second is also zero so that means the determinant of the three by three metric is very easy if you write the a cofactor expansion you only need one times a two by two so just one term so it's very easy okay so you can try that uh let's let's do a like a regular
5:25:11
problem so how about this one uh find
5:25:18
the general solution of final general solution of 2 y triple prime plus y double prime plus y equals 0 something like that so it's a third order differential equation ah now what we can do same argument we can look at the characteristic equation so the answer we can look at the characteristic equation is going to be 2 lambda cube plus lambda squared plus uh 1 equals 0 so this is what we get so it's a cubic equation and you can see that there are no four terms that means grouping doesn't work for
5:26:15
this problem what we can do we can use the factor theorem for that okay so ah so let f lambda equal to lambda cube lambda squared plus and if you are not familiar with how to find factors of the things like that please look at the factor theorem video and there i explained like many different type of problems like in a quality cubic uh quality like in order for a queen t quarter five and even like over eight nine ten like large ones also we basically use the factor theorem and
5:26:48
and then the long division of the synthetic division we normally we go for the center division we first use the factor theorem and then use the synthetic division that's how we normally do it so let's let's see how it works here after you get that now we can notice notice that and then let's say note that now that what we're going to do is we're going to try these values we can try uh lambda equals 0 plus 4 minus 1 and plus 4 minus 2 we're going to try those values
5:27:23
until you get a quadratic that means you only need to find one value because after that you can divide then you can get the quadratic so use the quadratic formula or factor so we just need one value to find so note that if negative 1 if you plug in the negative 1 what can happen 2 negative 1 cube negative 1 cube plus negative 1 squared plus 1 how much you get you get negative 2 plus 1 plus 1 which is 0. what does that mean by the factor theorem this means that so this means that lambda plus
5:28:10
1 y plus 1 because it is minus minus 1 because minus we plug in the minus 1 so this is a factor theorem so lambda plus one is a factor or in other words negative one is a zero so now what we can do we can use synthetic division to find the other one so use synthetic division
5:28:40
to find the other factor to find other factors so let's see how it works in this division we only use the coefficients so the coefficients are two of the original polynomial uh and then x square x cubed plus two and then one there's no lambda term so that means you'll put a 0 there and then 1. so those are the coefficients so we check negative 1 whether negative 1 is no what we are doing is we we want to get the quotient when you divide by lambda plus 1 so what we put here is the zero
5:29:25
corresponding 0 and you can see the lambda 1 lambda plus 1 means the 0 is negative 1. so that's the 0 so that's the one we're going to use there what we normally do is uh it's a very simple one we copy that number first and then multiply and add we're going to multiply negative 1 with 2 so you get negative 2 put it there and add secant negative 1 then multiply again with negative 1 and negative 1 so you get 1 and add so you get there just add multiply add and then negative 1 negative negative 1
5:29:56
with 1 so you get negative 1 and you get 0 that means good that means it's completely divided if you get a 0 that means it's completely divided so that's the quotient so we have the other factor right here 2 minus 1 1 those are the coefficients so that means the other factor of the factor is 2 lambda squared why there are three terms minus lambda plus 1. so we get that so that's the other factor now what we're going to do uh so we're
5:30:29
going to set that equal to 0 because we looking for that so we use you can see whether you can factor you can see that it's not easy to find factor so we're going to use the quadratic formula so let's use the quadratic formula so we have 2 lambda squared minus lambda plus 1 equals 0. lambda equal minus 1 is plus plus or minus 1 minus 4 a c square root over 2 a that means 4 so let's simplify you get plus 1 plus or minus you get 1 minus 8 so that means negative 7 square root
5:31:15
of 4 using complex numbers we can write this as 1 104 plus or minus square root 7 over 4 i so ah that is the alpha that's the beta so we have n beta now we can it's a complex so the complex conjugate
5:31:40
complex conjugate so complex conjugate means we can use the complex form so the three roots are so we can see thus the 3 roots are lambda 1 is negative 1 and the other one is 1 4 plus so minus square root 7 over 4 i ah and as before the alpha and beta so we write general solution so the general solution
5:32:13
so the general solution we can write which is y equal c 1 e to the negative t plus let's write the form first so what is the form actually uh because this there's a one real uh and then complex so that is c we can write this one as c 2 e to the alpha t cosine beta t plus c 3 e to the alpha t sine beta t and if you plug in that here you're going to get c 1 e to the minus t plus c 2 e to the uh what's the alpha 1 4 so 1 4 t cosine square root 7 over 4 t plus c 3 e to the 1
5:33:17
4 t sine square root 7 of 4 t so that is the solution of the third order differential equation so we use quadratic equation c 3 division factor theorem so it's like nice problem okay so this is the uh end of homogeneous equation now let's discuss how to solve the non homogeneous equation
5:33:48
so let's talk about the next very important type so uh this is the how to solve a non-homogeneous equation so this is topic 21 solving non-homogeneous equation solving and non-homogeneous
5:34:17
equations
5:34:22
so uh so we can start with a problem like this a y squared plus b y prime plus c y equal f x this is again a constant coefficient situation so with a constant coefficient how we do like when you get a problem like that we're going to split the problem uh we're going to consider the homogeneous part
5:34:57
and the non-homogeneous part separately like this
5:35:06
so from the homogeneous one we find a general solution so find the general solution general solution what we normally call the complementary solution so this has the format y c as a complementary solution c 1 y 1 plus c 2 y 2 c 1 y 1 c 2 white so that's the general solution and for the other one we find one solution so find a particular solution
5:35:48
find a particular solution we're going to call this yp uh so just one solution one solution just one solution
5:36:02
so we don't mean only one solution now what we can do after we know the particular this is what we call the complementary solution so this is called complementary solution and this is the particular solution
5:36:27
particular source and then we can have a general solution so the general solution is
5:36:35
general solution to the non-homogeneous is given by y equal we just add the 2 y c plus y p so you're going to get c 1 y 1 plus c 2 y 2 plus that one solution that you find so this is what we do we just need one solution and we add that to the general solution now the argument is you know that the non-homogeneous equation is a second order differential equation that means if you uh find the general solution you need two constants you can see these are the two
5:37:18
constants so everything looks good so because second order you need two unknown uh two arbitrary constants we have two arbitrary constraints c1 and c2 okay but yp's only one solution called the particular solution now how to do this in reality so what we do is we're going to use two techniques for that uh use use two methods two methods uh to find yp because we only know how to find yc so the two methods are uh first one is what we call the undetermined coefficient method undetermined
5:38:04
undetermined uh coefficient method
5:38:16
and we also use the variation of
5:38:22
parameters variation of a parameters method
5:38:30
variation parameters method so the problem with the undetermined coefficient is you can only solve three type of problems so problem with like a exponential function like that and then sine cosine and then the polynomials uh so polynomial those are the only three types that you can solve using the undetermined coefficient method so if you get something like secant you cannot do using this method uh like natural log you can do using this method so there's a problem so we need new
5:39:08
technique that's what we call the variation of paranoia's method uh if you have something like that okay so let's first talk about the undetermined coefficient method this is a very powerful technique uh so let's talk about that first undetermined coefficient method
5:39:33
undetermined coefficient
5:39:42
so this is like simple idea we kind of guess the solution and plug in and try to find those undetermined coefficients uh so again work for three cases only work for three cases only that's a major problem so we have a y double prime plus b y prime plus c y equal let's say if x x is the independent variable so it can be exponential function that means if we have like 5 e to the 3x we can handle that then we have the sine and cosine functions for example if you have 3 cosine 2x minus 4 sine x
5:40:39
we can do that and you can see that this is 2 this is 1 that's okay we can still do it and the third case is the polynomials so like a 2 x cubed minus x plus 1 we can also do that one so those are three types that we can do exponential sine cos and polynomials so what we normally do is uh the first step is guess the particular solution guess the form of pi p so that's the what we we're going to guess we guessed the form of yp uh so but there are certain criteria for that so let's talk about that
5:41:25
we have ah so guess
5:41:33
form of yp so we're gonna follow this table the first type if we have a number times e to the beta x beta is another number so your y p is a form of y p uh gonna be a e to the beta x the goal is to find b uh i'm sorry find a goal is to find a that's the first time the second type there are three cases it can be a something times cosine beta x something times sine beta x o let's say uh something times cosine beta x and another thing times another number times sine beta x so
5:42:33
here you see that the beta beta so in this case all these three has only one form all these three if you see any of them it's going to be simply a cosine beta x plus b sine beta x even if we have only one like just cosine sine we still need to have two the reason is when you take the derivative it's going to change sine to cosine so that's why we need to add both so that's the second type the third type is a polynomial type so it is like nth degree polynomial
5:43:19
and degree of polynomial so you're gonna guess the form as uh a say nth degree so it's a a n x n and then a n minus 1 x then minus 1 write all the terms and see until a a constant so write all the terms okay all term you cannot miss anything okay even though the actual function f x doesn't have that we have write all of them here so write all terms it's very important good so that's the kind of idea so ah so let's talk about some special cases so not so there are some special situations for
5:44:08
example if beta beta is a root so root of the characteristic equation characteristic equation of y c the complementary solution then then y p has to be then y p has to be a you have to add extra x in front so that's very important because if you do not add extra uh x in front if beta is a solution what will happen a e b x is a solution so it just disappears it become 0 so you don't get anything so that means you have to add an extra x okay together to get to get a
5:45:04
independent solution to get an independent uh solution it's very important you need that extra x there to get a independent that happened when beta is a root uh and then uh if b is a repeated root if let's say beta is a repeated root repeated root of the characteristic equation then you need to have another x so that means there will be a not just 1 x you have one more x or a x squared a b that's what happened if it's a repeater root you need to add one more to get independent solution ah
5:46:03
and then the next one is how about you're going to see a combination of the two or three so let's talk about that so the next one is if there's a combination if there is a combination
5:46:26
if there's a combination of them
5:46:32
write
5:46:37
write yp for each and combine write yp for each combine so for example let's look at some examples so let's say your f x looks like this 3 x minus 2 e 4 x like that so what we can do you can see that there are two there's a polynomial and an exponential function what we can do we're gonna have both in there so because of the first uh degree polynomial you need a term like ax plus b is the first degree polynomial that's for 3x plus because of the
5:47:23
exponential one you have c e 4x so we're gonna write for each and combine like that so you just write one expression combining both so that's the idea it's very simple uh the next one is how about we can see product and like all of them for example x e to the 4x a plus sine 3x so if you get something like that now the yp looks like this so we have x that means a polynomial so we write a x plus b for that and then we have e to the 4 x but for e to the 4 x the general term looks like
5:48:07
c e 4 x but there is something here because a and b already contained c as a factor so that means we are not going to write that c here so we don't write that we don't include that there because c already containing a and b so if you add a c that's extra unknown so that's gonna make it difficult so we are gonna write that so we're gonna remove that that means we without that uh we can write like this just uh a x plus b e to the 4 x
5:48:44
like that so that that argument always there if it is already inside as a product we don't need an extra one so so that's the first two so this can take care of x and the e to the 4 x now sign this sine this is one term but we know that you have sine and cosine both so that means this is c sine 3 x plus d cosine 3x because if you see one of them you need to write both okay this automatically comes everything good so that's the kind of idea and then also
5:49:18
let's say you also have extra term like minus cosine x so that means this also gonna introduce another two so that means if you if you have that then what's going to happen it is e again uh sine cosine it doesn't matter so write cosine and also the sine version f sine is always go as pairs okay so it's very important good uh so that means so this is go with that and this one go with that so that's the uh idea okay sorry it's clear now what happened uh
5:50:00
uh the other one that we just discussed before uh so let's say if you if you have a difference equation like this let's say you have y double prime minus y prime minus 2 equal ah 3 e 2 x so if you look at this problem so if you want to write the form for yp let's look at the it's always a good idea to look at the characteristic equation first if you look at the characteristic equation it is lambda squared minus lambda minus 2 equals 0 the factors are lambda
5:50:38
minus 2 and lambda plus 1. so this says that lambda equal 2 and minus 1 you notice that this exactly match with that so in that case what's going to happen uh your yp has to include an extra x so it is a x e to x the reason is uh lambda already appear in the characteristic equation because if you just have a e 2 x it's not going to work the reason is it's a solution since the solution once you plug into the left side it just gets 0. so you get 0 equal something
5:51:21
non-zero that's not possible you can have 0 equal to non-zero so because of that to avoid that we introduce that extra x to get an independent solution okay so to get an independent solution because we discussed that because if you multiply by x it's going to create a new independent solution okay so that's the idea uh so what we can do now is uh let's talk about the the steps to solve a problem like that so these are the steps to solve a problem using undetermined coefficient method so steps
5:52:05
so the first is the first step is always going to be fine why c the reason is then it's going to tell you whether the lambda is appeared in the characteristic equation so we find y c first at least the lambda values that's the first the second step is guessing why so you're going to guess
5:52:33
so guess yp you guess yp that's the second step third step uh you can take derivatives take enough our derivatives
5:52:55
what does that mean if it's a second order differential equation you need two derivatives it's a third order differential equation you need three derivatives like that as i say enough okay sufficient number of derivatives okay so it's clear and then what we're going to substitute into the equation so substitute
5:53:19
into the differential equation at the non-homogeneous one and then ah we can equate the coefficient of linearly dependent terms so number five equate coefficients of linearly independent a terms linearly independent terms you're going to equate according to linearly independent terms and then uh terms to find uh to find unknown coefficients let's see if an unknown coefficient find unknown coefficients the undetermined coefficients and then after we
5:54:17
find that uh write the general solution write the general solution
5:54:29
in the form
5:54:33
in the form y equal y c plus y p you can write that and then last step is we're going to find you can use initial conditions to find the c's appeared in yc okay use initial conditions that's the last step use initial conditions uh at the end
5:55:04
at the end to find a c1 and c2 so that's we do at the end that's the last step initial one we use at the end so okay so that's those are the kind of steps to solve a problem using the undetermined coefficient method so let's just do one example to cover all those things okay let's do one example okay so let's do this one so example
5:55:38
solve the initial value problem y double prime minus y prime minus 2y equal s 6 plus 8 times e to the negative 2 t minus 4 cosine 2 t with y 0 equals 0 y prime 0 equal 2 so this is a complete second order initial value problem and with the complicated uh non-homogeneous term let's go step by step the first step is find the characteristic equation okay so uh the characteristic uh equation of the homogeneous equation by double prime minus y prime minus 2 y equals 0
5:56:45
is simply lambda squared minus lambda minus 2 equals here the next step is factor it's very easy to factor so you're going to get lambda minus 2 lambda plus 1. this can give you the lambda values lambda equal minus 1 and 2 so we're gonna the yc so the yc equal c1 e to the negative t plus c2 e to the 2 t ok now you can see that that 2 and that one is different so that means you don't need that extra t for this one so we can write the y p that's why we need that first
5:57:32
to see whether that's the right form so right uh yp so yp is let's go term by term 6 6 is a constant let me even add an a for that and then that's the polynomial part and then we have 8 e exponential terms so that is b e negative 2 t and then we have sine cosine so it's a cosine so we're going to go with c cosine 2t and also you do have the sine term so that's the rule okay good whether you have two or one you still enter two uh the next step is take derivatives so
5:58:19
number three derivatives so you can take derivatives uh so we're going to get y p prime uh so that's a 0 so you get the exponential function when you exponential function when you take the derivative uh that the constant comes out that we know so that means it's going to be negative to be e to the negative 2t so that's what you get from that the constant comes out and then it's a cosine cosine when you take theta is going to be a negative sign also 2 comes out so it is negative 2
5:59:01
c sine 2 t and this is sine become cosine 2 also comes out so 2 d cosine 2 t so that's what you give me the first derivative now take one more derivative because the equation second order so y p double prime another negative rule comes out so it becomes 4 b e to the negative 2 t sine become cosine 2 comes out so negative 4 c cosine 2 t cosine we can sine so negative 4 d sine 2t that's what we get so we get all the derivatives now and yp so we can substitute that into
5:59:52
the equation so let's do that step 4 substitute
6:00:03
substitute into uh the equation so let's do that so y prime uh so y double prime so why did i y so we're gonna put the y double prime term first so if you want you can repeat the equation here uh sorry y double prime minus y prime minus 2 y equal f t that's the function we have so let's put that so y double prime terms first sorry is 4b e to the negative 2 t my negative 4 c cosine 2 t minus 4 d sine 2 t so that is the y prime term now we're going to go with the uh
6:00:52
first derivative term it's a minus uh you're going to get negative 2 b e to the negative 2 t negative 2 c sine 2 t plus 2 d cosine 2 t and then y term which is negative 2 times the y term which is a that's the longest term uh 2 e negative 2 t plus c cosine 2 t plus d sine 2 t which is equals to the f t which is 6 plus 8 minus 2 t minus 4 cosine 2 t good so that's the expression we get now what we're going to do we're going to equate the coefficients of linearly independent terms so what
6:01:54
are the linear independent terms you can see here you have a constant which is a linear independent term and then we have our exponential function which is linear independent we have sine which is linearly independent we have cosine which is also linearly independent so those are the four terms that we have here which are linearly independent so step five uh equate
6:02:24
uh equate coefficients of linearly independent terms so we start with the first term you start with the constant so let's start with the constant term uh what are the constant terms we have uh you're going to see a is a constant and that's the only term we have and then other side 6 so we're going to equate those two so we have negative 2 a equals 6 this directly going to give a a has to be negative 3 so we all know that a value
6:02:56
same thing let's go to uh exponential function next so e to the negative 2t so we equate the coefficient of that it was also in independent term so let's start with that so you can see we have that e to the negative t and then we have this one and we have that one and on the other side we have this so those are the four terms that we have so 3 on this side 3 on the other side so we have 4b and then we have negative negative positives 2 b and then we have negative 2b
6:03:39
ah good and then equal on the other side we have eight so how this simplifies you can see this and that get cancels so you will get 4b equal 8 in other words b equal to so we only know a we all know b now the only thing that we need to do is uh find the coefficient of sine and cosine so let's go with one of them so let's go with cosine 2t they are also linearly independent so if you go with that you have cosine term here you have cosine term there there
6:04:24
and then this those are the four terms so let's go with one by one so the first one is negative four 4c the next one is negative 2d and the next one is negative 2c those are all the terms on the left side equal the right side negative 4. see whether you can simplify this a little uh so you can see negative 4 c negative 6 c so negative 6 c minus 2 d equals negative 4 what we can do is we can divide this sum by negative 2. if you divide this sum by degree 2 you're going to get equation
6:05:08
3 c plus d equal 2. so i'm going to call this equation number one now what we can do we're going to create a similar equation for our sine 2t and solve the two so let's see what you're going to get for sine
6:05:26
for sine we have sine 2t we're going to get negative 4d and then plus 2c minus 2d equals 0 but there's no sine term inside this simplifies to negative 6d plus 2c equals 0 and if you divide this by okay i'm gonna write it here uh so this can give you uh negative 6d plus 2 c equals 0 if you divide by
6:06:15
2 you're going to get c minus 3 d equals zero i'm going to call this equation number two now what we're trying to do uh we can simply uh get rid of d term by so i'm gonna multiply the first equation by 3 and add to the second equation when you do that you're going to get 9c plus c minus plus 3d minus 3d which is equal to 6 and you can see these kids cancel out so you will get the value of c so 10 c equals 6 it says that c equals 6 over 10 in other words 3 over 5
6:07:08
and then we can use the second equation so d is one third of that so this says that d equal one third of c which is one fifth so we find the two values now we can write the general solution so thus uh the yp or the general solution yp is negative 3 plus 2 e to the negative 2 t plus 3 5 cosine two t plus one fifth sine two t so we get all the terms now what we're gonna do uh we got the general solution so number six the general solution so the general solution
6:08:04
is y equal y c plus y p ah so you can substitute is c 1 e to the minus t plus c 2 e to the 2 t and then minus 3 let's add them plus 2 e minus 2 t plus three fifth cosine two t my plus one fifth sine two t so in the exam like you know you're going to get a question like this this long but here i'm telling you like you can manage all of them together but you will get like one or two terms not like a lot like that okay now the last step is now we can use
6:08:51
the initial condition to find c1 and c2 so you can see that the only thing that we don't know is c1 and c2 to find that uh we can use the initial conditions so initial conditions we use at the end that's what you remember so the step seven are using the initial conditions we have
6:09:17
y 0 so 0 which is y 0 that means you plug in 0 for all the t's so you're gonna get uh what are you gonna get so let's write a separate equation for that yeah i'm gonna write it like that it's much easier like that uh using initial using initial condition y 0 equals 0 we have 0 equals c 1 plus c 2 minus 3 plus 2 e 0 plus 3 fifth cosine zero plus one fifth sine zero now we know that uh sine zero is zero cosine 0 is 1 e 0 is also 1 so this is going to give you an equation
6:10:13
so what what equation you're going to get you're going to get 0 equals c 1 plus c 2 ah c2 we have uh what we hear we have minus three uh and also we have three minus three and we have two here so it's a minus one plus three fifths that means minus two fifths that's what you get so i'm gonna call this equation number three because we already have one and two now the same thing let's try to use the other conditions we get one result here now first we taking derivatives taking uh derivatives first
6:11:01
are we gonna get y prime t equals to negative c 1 e to the negative t plus 2 c 2 e to the 2 t minus 0 minus 4 e negative 2 t minus 6 fifth
6:11:28
sine 2 t plus two fifth cosine two t and now what we can do we're gonna use the condition uh using y prime zero equal to we're gonna get this equals to negative c1 plus 2 c2 minus 4 and then negative 6 fifth sign is zero so that's gone plus two fifth cosine zero okay so if you plug in the values because uh again this is zero and this is one you're going to get uh if you move negative for the other side six equal negative c1 plus two c two plus two fifth
6:12:34
i'm gonna call this equation number four now what we're gonna do we're gonna add them once you add them this can cancel out you get c2 so let's do that and also you see this is negative this is positive they cancel out also so that's interesting so by 1 plus 2 we have 6 equal 3 c 2 in other words c 2 equal 2 and then you can use one of those equations so let's say c1 this is the first equation c1 equal ah negative c2 plus two fifth and we can plug in so it's negative two
6:13:24
plus two fifth equal negative eight fifth so we get the solution so therefore we have the solution y as negative 8 5 e to the negative t plus 2 e 2 t minus 3 plus 2 e negative 2 t and then 3 5 cosine 2 t plus 1 5 sine 2 t you can see that none of the terms are zero that's very interesting okay so that's the solution so it's just like a writing few terms but otherwise it's not that difficult but you want to make sure that you uh kind of follow the pattern against the proper uh
6:14:24
yp term and then after you get the general solution so you can see the most lindish part was to find c 1 c 2 otherwise the problem was not that difficult the finding c 1 c 2 takes time because of a lot of terms but if i say find a general solution it won't be a very difficult problem ok yeah let's expect something like that good so the next topic is the uh uh variation of parameters and the last that's the last one we're gonna talk
6:14:58
about in this uh this part of the review uh in the next one we're going to talk about the series solution laplace transform and the system of equations so the only thing we will discuss in this one is the variation of parameters but that's the that's the easy one because you are allowed to use the formula so i'm going to discuss the formula and the technique so let's talk about that so let's talk about the last topic the variational parameters method so this is 22 uh the of parameters variation
6:15:33
of parameters method we know that the undetermined coefficient method only works for three cases so the problem is what if you have something else uh so we can use a variation of parameters in those those cases so this is a little bit more powerful than the undetermined coefficient problem so why we use this one we use this one to find find yp the particular solution of y double prime plus p x y prime plus q x y equal f x we it's the same problem but it has to
6:16:22
be in this form so we write that standard form okay if it is not in this standard form you have to bring it to stanford first so these are the steps so let's go with the steps uh so the first one i find two linearly independent solutions find two linearly dependent
6:16:48
find two line independent solutions of the homogeneous equation y double prime plus e x y prime plus q x y so you have to solve this first uh so it can be constant coefficient or maybe uh coach euler so that's the first thing and then also uh the px and qx maybe no need not maybe a constants may be constants okay so that's the first you find ah two solution so let them be so linearly independent ah and then let them be let them be y1
6:17:41
and y2 so let's call those two solution of y and y2 those are the solutions of the complementary equation uh the homogeneous equation now what we're going to do we're going to guess the solution so guess so it's the extension of the reduction order guess yp of the form of the form this y p of the form y p equal u 1 y 1 plus u 2 y 2 so for the non-homogeneous we have one one other one other term if you have the homogeneous it's only
6:18:21
u and y one but when you have non-homogeneous we add one more term so the goal is to uh find this u1 and u2 so that's the goal how to find this u1 and u2 so we do a very similar thing we take the derivatives and plug in to the equation and ah get two uh two equations to solve for u1 and u2 so you can see there we are u1 and u2 are unknown functions uh
6:18:58
where u1 and u2 are unknown functions so we need to find those two okay so the goal is to find u1 and u2 uh so what what's going to happen when you plug in you're going to get uh somebody skip the uh other parts so we're going to get end up with these two equation so we're going to solve the two equations solve the two equations solve the two equations
6:19:32
so the two equations to find q1 and u2 but two equations these two ah it is u 1 prime y 1 plus u 2 prime y 2 equals 0 that's one equation the other equation is u 1 prime y 1 prime as well plus u 2 prime y 2 prime equal the function f x so we can solve these two equations uh that's always the case you're going to end up with these two equations so the goal is to solve these two equations and find u and u2 and then after you
6:20:19
find u1 u2 you're going to plug in here you get yp so that's how you find the solution so let's use this method uh to solve a problem like that you cannot do using the undetermined coefficient method so let's see uh this problem example find the general solution of find the general solution of general solution of y double prime plus y equal 10x so if you have this equation you cannot do uh you can solve this using undetermined
6:21:03
coefficient because 10 doesn't work so the uh problem so what we're trying to do is just try to find the two solutions to linearly independent solution of the homogeneous equation so that we can do very easily uh the the characteristic equation of y double prime plus y equals 0 is lambda squared plus 1 equals 0. uh so lambda equal plus or minus i which you can write as 0 plus or minus i uh so thus using the constant coefficient argument
6:21:51
uh you don't get the exponential term you only get the sine cosine terms so that means yc has the form c1 cosine x plus c to a sine x now what we can do we're going to pick these two functions as y1 and y2 the order doesn't matter so let uh y1 equal cosine x and y2 equals sine x so let's pick those two functions ah now what we can do is now we're gonna uh so suppose assume by p of the form u1 y1 plus u2 y2 in this case it is u1 cosine x plus u2 sine x now we can take derivatives then
6:22:59
uh from the equations the two equations we're gonna get u1 prime cosine x plus u2 prime sine x equals 0 as the first equation the second equation says u 1 prime take the derivative so once you take the derivative of cosine it is negative sine so it is negative sine x plus u 2 prime cosine x greater than the derivative which is equals to whatever on the other side so in this case tan x so those are the two equations that we need to solve one and two now we're gonna think of how to solve
6:23:39
what we're gonna use the elimination type of argument uh you can see you can eliminate you one term easily what we can do we're gonna multiply by the missing term and add them it's gonna goes away so that's the kind of idea so let's do that so simply by we're going to multiply the first equation by sign the second equation by cosine because those are the missing term we'll multiply these two and add them so we're going to go away so by 1 time sine x
6:24:09
plus 2 times cosine x if you do that when you add them just going to go away on the other side what will happen are you going to get u 2 prime sine squared x plus u2 prime cosine squared x equal so the bottom one you multiply by cosine so it is tan x times cosine x uh which is what is the time mean sine cosine is sine x so which is equal to sine x if you simplify and here what's going to happen uh you can see this is 1 so you're going to get simply u
6:24:54
2 prime this is sine squared x plus cosine squared x which is equal to sine x but this is 1 so this implies u 2 prime equal sine x sorry now what you're going to do you take the integral you get u2 so if you integrate so u2 is simply the integral of u 2 prime dx so that means uh you just integrate that so that is sine x dx that means negative cosine x so that is u2 now by one by one you have u1 prime cosine x plus u 2 prime sine x so if you plug in you're going to get u1 prime cosine x plus
6:25:55
sine x that's the derivative of negative cosine times sine x which is equals to zero that's why it says now if you look at the last equation the last part which says that u1 prime is sine square so negative sine squared divided by cosine x that's what you get now what we can do think of how to integrate this one so what we can do is we can write the top as uh cosine squared x minus 1 over cosine x is you can split as cosine x minus 1 over cosine means secant x so that means it's very easy to
6:26:43
integrate so u1 is the integral of u1 prime dx uh so let's integrate the two separately so cosine x dx minus the secant x dx and we know the cos cosine cosine means the sine x that's the easy part and then
6:27:15
so we have minus we know that the secant secant means natural log secant 10 secant x plus 10 x that's the formula from calc 2 uh we are going to add c so
6:27:34
okay so that's a that's a formula you know the secant uh integral of secant is c uh ln secant 10 that's a formula okay so now we have that so that means we can write the solution now uh so the yp is simply u1 y1 plus u2 y2 which is equals to cosine x times sine x minus uh
6:28:09
minus ln secant x plus 10x plus u2 uh u2 means negative cosine so it is sine x times negative cosine now you can see that this and that get cancelled so you only get the last term uh so that means the yp is negative cosine x ln secant x plus tan x so the general solution
6:28:58
the general solution is y equal yc plus pi p c1 cosine x plus c2 psi x minus cosine x ln secant x plus tan x okay so that is the general solution which has two constants c1 and c2 good so that's the kind of idea how to use the variation parameters method uh you're going to write the two formula two equations for u1 and u2 and this ah use the elimination method to remove one of them and solve for a one and they use one of those equations
6:29:45
again to solve for the other one uh the variation of parameters also has a formula so let's discuss that and you are allowed to use that in an exam so let's talk about the variation parameters formula so uh the formula
6:30:06
uh formula for variation parameters variation of pyramidal's method
6:30:23
so let's talk about that ah again the same idea uh let's say two linearly independent solutions to independent solutions of y double prime plus px y prime plus qx y equals 0 a r let's say y 1 and y 2 that's always the first step we find two linear independent solutions of the standard equation and then the particular solution is given by a particular solution particular a solution of y double prime plus p x y prime plus q x y equal f x
6:31:34
is by by p equal it is simply u 2 y 2 y 1 f x over the wranskian y 1 y 2 automatas minus y 1 y 2 f x over the same ransom y 1 y 2 d x so you can remember this formula and use uh
6:32:23
when there is a variation of parameters type of problem because there are a lot of problem that you can use the undetermined coefficient method but this works only problem is whether this integral can be done because sometimes what happened this is not gonna work okay sometimes it's even difficult so this is simply the wrong scheme now you know that why uh the solution has been independent otherwise the drawn scale can be zero so we make sure that it's not equal to zero
6:32:51
because we divide by that okay let's just do one example from that just to see how it works um and then that is the end of the review too so let's do this problem so uh so let's say a use use uh the variation of parameters method variation of parameters method
6:33:20
method
6:33:24
to find the general solution enough to find the general solution of
6:33:34
solution of this equation y double prime minus 2 y prime plus y equal e x over x to the fourth you can see that uh this c this one you can cannot do using the integra unless undetermined coefficient method see the format because there's something in the denominator so then you can't do it so when you take the derivatives you're gonna powers get more and more and more so it's not gonna work okay so let's use the uh variation of parameters so the answer so first we can look at
6:34:09
the characteristic equation homogeneous equation so the characteristic equation of the homogeneous equation
6:34:19
is lambda squared minus 2 lambda plus 1 equals 0
6:34:28
the zeros are zeros ah you can see it is a lambda minus 1 squared so that means lambda equal 1 1 that's a repeated solution so does the y c is c 1 plus c 2 x e x or in other words c 1 e x plus c 2 x e x so that means those two are the y 1 and y two the two solutions y one by two okay so uh so take a peak or select y one equal e x and y 2 equal x e x as to independent solutions independent solutions ah now the wronskian
6:35:46
w y1 y2 so y1 first y1 y2 y1 prime y2 prime so if you plug in you get ex x x e x that means e x plus x e x if you multiply through you get e x times e x plus x e x minus x e
6:36:21
x times e x so you can see uh this and that get cancel out so you get e to the 2x which is not t equal to zero perfect okay so that's the wrong scale now the solution so the particular solution is the particular solution
6:36:48
by the formula it is y p equal y 2 y 1 fx over ramskin y1 y2 in that order minus y1 by 2 fx over w y one y two in that order so the same same thing no changes like with the negative sign in between so let's plug in so if you plug in you're gonna get x e x e x that's y1 and we have ex over x to the fourth power over wranskian which is e to the 2x dx minus y1 that is ex we have what we have x e x again e x over x to the fourth over e to the 2x dx and you can see
6:37:56
there are a lot of cancellation going on e uh this cancels with that the same thing happening on the side so you simply end up getting x e x the integral of 1 over x to the fourth dx minus e x times the integral of one of x to the cube
6:38:26
so these are very easy to integrate so if you integrate them you get x e x times x to the negative 3 over negative 3 negative e x times x to the negative 2 over negative 2 if you simplify
6:38:48
ah you get what you get negative ex over 3 x squared plus e x over 2x squared if you add them you get e x over 6 x squared okay that's the yp now what you do how you write the general solution so the uh the general solution y equal the sum of the 2 y c plus y p y c is what which is c 1 e x plus c 2 x e x plus e x over 6 x squared so that is the general solution using the formula so it's not long uh but you need to simplify the integral this is the end of the review too so
6:39:55
next we're gonna talk about uh review in in review three we're gonna talk about the series solution at least two examples and the laplace transform and the eigenvectors method
6:40:20
you this is the part three of the final exam review of the differential equation class uh in this review uh we're gonna talk about uh three topics uh the first one is the series solution method uh we will discuss two problems first order and second order and then we're going to go to laplace transform method so we will go with the definition and then how to solve a differential equation and the convolution and the heavy side function or the switching functions and then the direct delta functions
6:40:52
uh and then uh finally we're gonna talk about the uh uh solving a system of ode so we will go through the three techniques uh the elimination method uh and then the laplace transform method and the eigenvectors method so let's start with the series solution method uh sorry is number 23. the series solution method series our solution a method
6:41:29
so this is a kind of a last resort type of situation because there are a lot of differential equations you cannot find closed form solutions so in that case what we're going to do we're going to assume a series solution we assume there's a solution given as a power series and then we try to find the coefficients of that and once you know the coefficients you pretty much know the solution so that's the kind of idea so uh we normally deal with secondary differential equation in this
6:41:58
course or uh mainly so we have a equation like this so a x uh let's say y double prime uh plus uh plus bx y prime plus c x y uh equals zero so let's say the homogeneous equation so what we're going to do we're going to assume so the first step we're going to assume assume
6:42:35
assume a solution of the form assume a solution of the form of the form of y equal sigma n going from zero to infinity c n x so we assume that the solution can be written in this form and then after we do that we're gonna assumption and then we can do we're gonna take the derivatives of this one so we're gonna take derivatives so depending on the differential equation with the first order you need one derivative if it is second order you can need two derivatives so take derivatives
6:43:12
uh that's the uh second step uh and uh plug in so we take the derivatives and plug into the equation and plug in ah to the um equation so we can take derivatives of the series and then plug into the equations so we have two steps so take the derivative and plug into the equation third one is after you find the equation we try to find a relation uh between the coefficients so we can try to find the relation among the coefficients okay so find a relation final relation um among coefficients
6:43:57
okay for a relation we can say between a coefficients
6:44:10
so it will be clear what we mean by that because sometimes you can express uh like what we try and do we can try to find a solution for c n plus 1 or c n in terms of other coefficients okay that's a relation we find a relation between the coefficients and then what we can do we can evaluate uh uh few a coefficient so uh after that we're going to find uh c naught we can probably have it so then we're gonna find the rest so we're gonna find we're gonna calculate
6:44:40
few and if possible if possible that we don't know if possible ah we can say uh if possible uh final formula for a c n or c n plus one normally c n plus one so we try to find a formula but that's sometimes it's impossible like i mean you can but it's gonna be like very complicated after that what you're gonna do we're gonna plug in them and then write the solution and then after we do that then we could write the solution uh write y which is sigma n going from zero to infinity we're
6:45:23
going to plug in here so c and x n so like you know so if you look at expanded versions like that so we can just plug in for the coefficients because that's what we found earlier so we're gonna plug in here so that's the kind of uh point so we do that and sometimes we can see a pattern and find a clause form solution but that's not always the case so find a closed form solution
6:45:54
ah again if possible so that's not always guaranteed okay these are the kind of steps uh to do a problem so let's look at a problem and see how it gonna work so we can start with the first order uh equation so let's say uh for example find a serious solution find a series solution find a series solution of financial distribution of pi prime plus 3y equals 0. so we're going to try to kind of find a solution for that y prime plus 3y y prime plus 3y 0. again we're just going to follow the
6:46:39
steps so the steps are going to assume
6:46:48
so assume y equal a series solution so y is given as a series so that's the assumption we assume that now what we can do then we can take the derivative so you can see that when you take the derivatives the first term is a constant so when you take the derivative what will happen the first term disappear because a constant so that means ah we're going to start with y prime which actually start with n equal 1 because the constant term goes away so we just take the derivatives if
6:47:25
it has a derivative x n that means n c n x you subtract one so that's the derivative and you can see that the important thing start with one because the constant term disappear so if you take one more derivative it's going to start with n equal to because the now the first term is a constant so if we take the second derivative it's gonna go away so that's that's something we need to remember okay good now what we can do substitute so we can substitute
6:47:57
so once you substitute what will happen is y prime plus 3y equals 0 so that means you're going to get y prime that means n equal 1 to infinity in c n x n minus 1 plus 3 times y y means in going from zero to infinity c n x n which is equal to zero on the other side okay so that's so we plug it so that step is over we plug it take the derivatives and then plug it now what we're going to do you can see that we check what terms we're going to get uh
6:48:39
so for example let's plug in the first value in the index so you can see for the first series the first value is n equals zero so you play an equals zero i'm sorry first value is n equal one if you play n equal one you get x to the one minus one with x to the zero so you get x to the zero in the first one second one you can see the index value is n equals 0 when you plug in n equals 0 you get x 0. so that means that's a good one that means you get exact same number in both
6:49:07
but if you do not get the exact same number we need to write the uncommon terms first we will write uncommon terms and then start writing the common terms so that's we're going to see in the next example but here you can see everything start with x0 so everything looks great now what we can do we're going to use the cs4 method i remember the cso method so we're going to use the season method to reindex i'm going to write it here so next what we do i use the seesaw method so cso method to reindex
6:49:51
so what's the reason it's a method that i introduced so you can find them in other books okay so you can look at the video about iso method it's there so what we're gonna do uh is very simple because you can see the second uh second sum it is x n we try to get x n in the other one as well how to get x n you can see to get x n you have to increase the index by one so you can increase this one by one what we do we increase inside decrease outside that's what we're going to see
6:50:18
so because if one goes up the other goes down so that's the kind of idea so it's like a seesaw so this side goes up then this side goes down so that this stays the same you see that the middle never change but one goes up one goes down so that's the kind of um idea here one when one goes up the other goes down so that's what we're going to see so method so it's very simple i is going to make the calculation much easier so if you use that so outside you can increase by
6:50:47
decrease by 1 that means your sigma can start with 0 because you go down by one and inside you can increase everything by one that means n become n plus one c n becomes c n plus one x is n minus one if you add 1 it will become xa perfect and then here i am going to bring the 3 inside so this is 3 c n x n now you can see that both have the exact same index exact same exact same power x n and exact same index that means you can combine
6:51:27
so we'll combine them into one so we could write this one as sigma and going from zero to infinity n plus 1 c n plus 1 i i'm not going to write x in term i'll leave it and then from the other one you get 3 c n now we can write the common x and outside so now what this is this says that the whole series equal to zero the only way this can happen is this term has to be zero because there's no other way that the whole series is zero so that means this implies that this
6:52:01
term the coefficient is 0 and that's exactly what we need so let's write that so we're going to get that n plus 1 c n plus 1 plus 3 c n equals 0 for the index signal index also starting with you can see 0 1 2 like that so it's always coming from the sigma whatever the value you can put it there ah now we can solve for c n plus 1 so c n plus 1 is if you move the other side is negative 3 over n plus 1 times c n so that is the relation that is actually
6:52:45
the uh relationship between the coefficient okay so you find a relation again this is only valid for in going from 1 2 3 like that it's only valid for those now what you can do just calculate a few three values okay so we're gonna start with this first value n equals zero so if n equals zero you're gonna get c one let's plug in just plug n equal zero so you're gonna get negative uh three one since it's negative 3 c naught that's what you get now we're going to plug in equal 1 by an
6:53:17
equal one you're going to get c 2 is become minus if you play n equal 1 you get 3 halves c c1 but we already know what is c1 so let's plug in the c one value when you plug the c one value what happen is you get negative three you multiply another negative three that means negative three squared you release it like that and then the bottom you get 2 c naught so that's what you get for the second coefficient and then go with n equal 3 so keep writing and until you see a pattern
6:53:51
so if you play n equal 3 what happens is it's still negative 3 plugging 3 this become ah so if when you plug in 1 you get 2 when you plug in oh actually i missed the value uh this n equals 2 for n equal to then you get that and then ah become 3 a c 2 now we know c 2 so we can find the c 2 value here so when you probably see 2 value because he took up from the previous step you can see you get another negative 3 so i'm going to write it like that over
6:54:25
3 times 2 here c naught so and then you can see that what happened every time you're going to multiply by uh the same index like that you just divide by that so that term comes extra that's gonna you add it to the previous one so that's what you see so that means uh if you want to keep going like that so at the nth step what's going to happen you're going to get c n so let's say minus one step if you go in minus one step you're going to get a c n equal minus 3 over n
6:55:07
here it is c n minus 1 that you can simplify you can substitute so what will happen when you substitute like that you can see that it is negative 3 so you can see that when you have 1 it is negative 3 only one of them when you have 2 there are 2 and you have 3 3 so that means in this case you're going to get n of n in the bottom you can see that that also that number exactly was the bottom number so that means in the bottom you're going to get n times n minus 1 times like that that's simply
6:55:39
the factorial so times c naught so if you go back what you have gained is negative uh so if you go back and you can feel this now because you see the pattern now so it is negative 3 squared factorial 2 c naught this become negative 3 cube over factorial 3 now this is negative 3 to the n over factorial n c naught so that's what you get so that is the uh general coefficient now what we're going to do we're going to plug in that in the equation so we found the coefficient
6:56:18
so let's plug it so therefore then you can see y which is sigma n going from 0 to infinity c n x n so if you plug in for c n you're gonna get um it is negative three to the n power of a uh factorial in uh and then i can write the c not outside so x n i would rather see not outside let's kind of uh make it simple so it is equal c not i write c dot outside is common so what is going on here so you can see this is simply c naught times n going from 0 to infinity you can write this one as
6:57:03
negative 3 x to the nth power if you combine if you now what is this so you can see this is simply the uh taylor series of e to the negative ah 3x so this is simply c naught e to the negative 3 x as expected because we know that this differential equation is one of the four fundamental equations we already know the solution it is c naught e to the k x uh in this case k is negative 3 that's exactly so we exactly get that so if you forgot what i'm writing here we know this
6:57:40
formula uh so e to the x is sigma n going from 0 to infinity x n over factorial in and you can see this is exactly what we have here here x is negative 3 x so that means you get negative 3x like that so so so this is the solution of the differential equation and you can see that's not that difficult after you find the relation try to evaluate few and see whether you can find a pattern and if you can find a pattern you can nicely simplify like that so uh yeah so that's about that so i'm
6:58:17
gonna do one more example uh from little like a higher like second order differential equation so so let's do a little complicated one just to see what we can do but in the exam the question is going to be reasonable so you will get a reasonable question but here i'm going to do a like question just to see what's uh what's going to happen okay so this is like a three term differential equation three term good so let's do that so the next example again uh find a serious solution
6:58:49
find a series uh solution of
6:58:58
find a serious solution of x minus 1 y double prime minus x y prime plus y equals 0 so that is the differential equation we're going to solve the same idea but this is second order so it doesn't really matter so but only thing is we need to take two derivatives so the answer so we're gonna say assume
6:59:27
uh a solution of the form assume a solution
6:59:34
of the form of y equal sigma n going from zero to infinity c n x n this step is always the same now we take the derivative as we discussed once you take the derivative you can see you can start with n equal 1 because the first term goes away so you get n c n x n minus 1. take one more derivative so you're going to get the new first term i'm going to go sorry because that's a constant so you're going to start with n equal to and then you have one more so it's n in minus 1
7:00:12
c n not change x to the n minus 2 now okay so there are two things to focus on this and that here this and that uh good so those are derivatives now what we can do we can plug so substitute into the equation so substitute substitute ah into the equation so if you substitute what we get uh we have x minus 1 that's a term and then y double prime so that means the sigma n going from 2 to infinity and n minus 1 c n x n minus 2 then minus x the first derivative which
7:01:02
can start with n going from 1 to infinity ah 1 to infinity n c n x n minus 1 plus y which is sigma n going from 0 to infinity c and x and so we write that so next step is we try to bring this inside before we change the index okay we're going to multiply through so let's multiply through uh if you multiply through what will happen uh you can see there's x term once you multiply by x term uh inside it's gonna be x to the n minus one and then there's a minus one in front
7:01:44
the minus one multiplied by this so we can start with that so that means the first term gonna be now that's the highest one so we're going to go with that so it is minus sign go inside n n minus 1 c n x n minus 2 because i multiply by the minus minus 1 and then let's start multiply with x when you multiply with x you get ah is still n going from 2 to infinity changes with the index n n minus 1 c n x when you multiply by x is because n minus 1 okay because you multiply by x term and then
7:02:24
from the next one we can bring the x inside uh so if you bring the x inside it's going to be sigma n going from 1 to infinity n c n x n why because you multiply by extra x so in minus 1 become n and then the last term was the last term uh sigma n going from zero to infinity c n x and equal zero so you get ah now so so we substitute and bring the x outside x term inside that's the next one now what we're going to do we're going to see whether what are the starting
7:03:08
terms so that's what we check so let's check the starting term if i n equal to the starting term is x to the zero if you plug n equal to starting term is x to the one okay so that's slightly different again in the next one if you play n equal one starting term is x with the one the last one x to the zero so that means there are uncommon terms because uh in the first one the last one there's x zero term that is uncommon because the other series the other two series does not
7:03:40
have that term so what we're going to do we're going to write the uncommon terms first so let's try the two uncommon terms there's a one uncommon well after that if you start with the next index that means n equal 3 you can see that you get x1 in the last one also if you start with n equal 1 you get x to the 1 and everything have x1 x1 x1 x1 on all four so that means they have common common index so we only write the uncommon index so uncommon terms
7:04:09
what are the uncommon terms you get uncommon term of the first one when you by n equal 2 so that means it is negative 2 2 minus 1 c 2 x to the 0 so that is the uncommon term from the last one find the last term you also get uncommon term when you apply an equal zero so we write that and then what we can do we gonna write the rest of the terms using the proper index so in the first one now you can see that n equal to we already plug in so we apply n equal
7:04:41
3 now okay because we the first one we already wrote so minus n n minus 1 c n x n minus 2 no changes just the index just the index the next term no change
7:04:59
because only for the uncommon term so no changes here yeah there's a lot of writing i don't know even in a space here and then i'm gonna write here negative ah no change because still this is a common term so we can write n c and uh x to the n from the last one it has to start with n equal not 0 but 1 because we already wrote the first term there so that is what we do next so we write we change the indexes only for those uncommon terms now what we're going to
7:05:35
do we're going to use the cs4 method now we're going to re-index now we can re-index and see whether we can bring everything into one common index that's very easy to do so let's look at the terms now you can see we need to bring it to xn if possible so how to do that we gonna uh bring this one by up by 1 this one down by 1 so then it's going to be x n power x n here you can see you want to bring this one up by 2 so this one down by 2
7:06:11
ah and then everything else in exchange so fine so we only need to change the index of 2 just 2 series so let's do that so we're going to get after we reindex so we're going to get so we can start with the uh
7:06:34
we will start with the with the first one and then uh only the the first two and then we have uh the other two we can combine them also because you can see the last two you can actually combine them into one uh so let's do that so we're gonna get
7:06:53
so we have this now so we have negative 2 c 2 plus c naught we have that and then we're going to use the c so method for indexing we're going to bring it down by 1 down by 2 down by 2 so it's becoming 1 because we had n equal 3 before it become 1 infinity not change now inside you can increase everything by 2 so you add 2 outside you subtract 2 inside you add 2 you can become n plus 2. n minus 1 become n plus 1 see n become n plus 2 x n minus 1 become x n plus let's go to the next
7:07:37
ah in the next term ah we need to bring it ah down by 1 so it's become n going from 1 to infinity so you add 1 becoming plus 1 and then we have n minus 1 that become n this is c become n plus 1 ah and then we have x n minus 1 that become x n and ah what we can do with the last two we can actually combine them into one because you can see they are the same so we can combine them so we have uh so if you want we can just leave it because just to keep the argument
7:08:15
so behind going from one to infinity uh n c and x n and the last term is n going from 0 to not 0. so again 1 to infinity so 1 to infinity we have c and x n equals 0. now what do you notice you notice that everything has the exact same index in one and one n one and one also everything has the exact same power uh so what does that mean you can combine them so we're gonna combine them so let's combine them so once you combine you will get negative 2 c 2 plus c naught now let's combine
7:08:56
let's write the common index outside so the common index is n going from 1 to infinity that's the common index and let's put those t those terms together from the first one you're going to get my negative n plus 2 n plus 1 c n plus 2 that's the first one and from the second one you're gonna get plus uh n n plus 1 c n plus 1 and if you combine the next two you can see uh one has a negative n c n the other one has c n so you can see you can write this one as uh uh
7:09:41
if you order the plus sign it is 1 minus n c n you can add it like that ah because uh one has a negative n c and the other one is c n so we can write like that and then write the x n power outside which is whole thing equals to zero now again if this whole thing equals zero what we're going to do we're going to equate the coefficients of each term each linearly independent term so you can see that this implies that the first thing implies that is this has to be 0 other thing is that this term
7:10:22
has to be 0 so that's what you get so let's put those two together let's set them equal to zero so if you said the first one equal to zero uh so you're going to get you can solve for c2 you can see that the c2 equal one half c naught that's what you get if you set this equal to zero and solve for c 2 you get that and from the second one if you solve for a c n plus 2 so i'm going to solve it here so if you solve for c n plus 2 just set this equal to 0 and solve for c n plus 2
7:10:58
so you're going to get simply n plus 2 equal you can see n plus 1 term get cancelled from the first two terms so you get n over n plus 2 c n plus 1 minus and then you're going to get n minus 1 i put the minus 1 outside it's easy and then we have here we have
7:11:28
here we have uh n plus 1 over n plus 2 so n plus 1 and plus 2 cn so that's what you get okay and this is valid for n going from what is the index value 1 so starting one so 1 2 3 you can see that's the exact same indexing value here see the system okay so that's a relation so that means we know the uh we know the relation though what we can do is we can calculate fuel we can start start plugging the values and then try to calculate a few of the uh terms so let's do that okay
7:12:06
so after that after you get the relation we're going to start calculating values so we can start with the first indexing value n equal 1 if you write n equal 1 are you going to get c 3 so let's plug in see if i in equal 1 what will happen is a 1 and then over n plus 2 that means 3 a c n plus 1 that means c 2 or minus if you put n equal 1 what will happen n minus 1 becomes 0 so you get 0 times c c 1 0 times a 0 times c 1 that means c 1 d disappear 0 times okay
7:12:47
so that's what you get um so in other words now what we're gonna do we're gonna substitute for c2 because c2 we already know from the previous step see c2 equals one half c know so let's plug in there since plug in there you get 1 over 3 1 over 2 c naught in other words you can see that later it is factorial 3 c naught factorial 3 means what what is factoring 3 factorial 3 means 3 times 2 times one so you're gonna see that gonna appear here okay but if you don't see it
7:13:20
that's okay when you write few more times you're gonna see that now let's try to calculate the next term so we're gonna calculate n equal two so by n equal two this becomes c see 4. so let's plug in n equal 2 in the equation it's a 2 over nope it's a 2 so it is 2 over ah so when you plug in 2 it becomes 4 c3 actually i don't want to write that large so this 2 over and so it is 2 over 4 c3 and then uh minus if n equal 2 that means for 1 over what do you get here when you play
7:14:05
n equal 2 to the equation you're going to get 3 times 4 times 3 c2 that's what you get now what we can do we're going to plug into c2 and c3 both of them so let's plug in so you plug in what do you get from the first one you get 2 over 4 c3 what is c3 it is factorial 3 c naught minus 1 over 4 times 3 plug into c2 c2 is one half c naught now you can see that this is they have the exact same denominator that means they are the exact same factorial 4 and then in the top it is 2 minus 1 2
7:14:48
minus 1 means 1 so that means it's nicely simplifies to factorial 4 c naught that's interesting the same thing n equal 3 let's choose one more then you can see it's clear so you're going to get 3 over 5 c 4 minus 1 over divide n equal 3 so it's going to be 5 times 4 c 3 so let's plug in here so you get 3 4 3 5 c four means factorial four c naught minus uh five four c three c three means factorial 3 c naught the same argument you can see the denominator is simply factorial 5
7:15:41
now and the top is 2 minus 1 again so that means
7:15:49
uh is that right uh
7:15:55
so uh yeah actually there's there's some typos here so it is not two this is three and also this is not 1 anymore so this is 3 so 305 and then uh here it is you plug n equal 3 this becomes 2 we have 2 or 2 divided by 5 4 with n minus 1 and you can see now it's again simplifies very well same denominator the top is 3 minus 2 that means 1 so it's going to be 1 over factorial 5 c naught so that means you can guess what's going to happen with 4 c n cn cn has to be 1 over n factorial
7:16:34
c naught so in general so that's why so once you write few terms you're going to see a pattern okay so try to find the you just write a few terms until you see a pattern oh if it's a difficult problem what's gonna happen uh the problem will ask you to write just the first five terms of like four terms five terms like that if it's a difficult one but if it is like like this problem very nice you can solve them you can simply go forward and find it okay so depend on
7:17:01
the problem it's a difficult one just like few terms is enough okay so let's now you can see in general okay in general you can write cn which is 1 over factorial n c naught which is valid and going from uh so what's the starting value uh we're going to start with n equal 3 so right yeah because we apply an equal one but we start with c3 so that means starting with three so three four five like that but uh we also know that and we know that c2
7:17:42
is one half c naught see fortunately that also the exact same format so that means we can simply write this one as c n is factorial ah phi factorial in c naught for n going from 2 starting with 2 not just 3 because it has the exact same format you see this is the exact same format good so that means now we know the general form so let's plug in okay so we can substitute that in the equation so substitute so when you substitute is y ah which is c ah so let's write the formula so this
7:18:25
is the sigma notation so we can do we're gonna plug in that so in going from zero to infinity c and x n so which is c naught plus c one x plus c 2 x squared like that so let's plug in here so when you plug in what do you get uh so you're going to get we write c naught we leave c 1 we don't get from here so we can leave that and then starting with ah c2 we have a formula so it is 1 over factorial 2 x squared 1 over factorial 3 x cube like that ah actually uh and also
7:19:11
we forget the c naught so let's write that so i'm gonna yeah you can put the top so this is c naught there's a c naught term everywhere you see not times ok so you can see so that is the case now what i am going to do ah you may or may not see a pattern here the only thing missing here is the c not term for the x to the first power so what we're gonna do is we can write like this now we can write this one as uh i'm gonna add and subtract that
7:19:49
so uh so what we can do i mean add and subtract so we can write c 1 minus c naught x that's the first term x 1 term and from the rest i'm going to add here c naught coming here that's that c naught term so i'm gonna write the rest so here we can add the c naught term so this is a c naught x term that's the one i add so this is subtract i so add that substitute now write the rest just write the wrist so what's the rest c naught over 2 factorial x squared c naught over 3 factorial x cube like that
7:20:29
and you can see from the rest so this c one c naught it's just a constant i'm gonna say it's a big a c one plus what's gonna happen here here we have this c naught uh term so you can pull the c naught out so if you pull the c naught out uh it's just like a just a term uh so what you get you get one plus x plus one 2 factorial x squared plus 3 factorial x cubed plus like that this is simply the e to the x so that means you're going to get this c
7:21:09
1 is not the c 1 the c 1 is simply equal to c one minus c naught so i'm gonna i just it's just a number that number i call c one uh the big c one so we're gonna get uh uh c one x plus c naught e to the x so that is the solution of the differential equation it's very well simplified and sometimes this happen you know you can do that but not always so this is the kind of thing uh but in a final exam in exam this may not be possible unless it is like a
7:21:47
like a short problem uh the first problem is okay but you know you will get a manageable problem but this is like pretty long a little long um unless you can do it really really fast but in exam you don't expect that uh because you know you have a lot of things to do so but what i want to show you how it works okay this is how it works sometimes it works very well uh so you will get something manageable okay not like this long so you can look
7:22:14
at the problem that you get for exam one thing okay good so so that is the power series method and you can like do any problem like that so it's not difficult but you can see it's like a simple thing so next we're gonna do is the we're gonna talk about laplace transform method
7:22:33
uh so let's talk about the laplace transform method so it's a number 24 the laplace transform method a plus a transform a method
7:22:53
so this is again a very powerful technique you can do a lot of problems using this technique because we know that we get stuck uh when you try to find the solution with undetermined coefficient measure because there are only three three uh problems that we can do three type of problems that we can do and then also if you try with the variation parameters uh you get an integral that may be impossible to um like get the clause form solution then we have this new technique what we call
7:23:18
the laplace transform method so let's try to uh look at the definition of laplace transform so the definition let's say you have a function so uh for a piecewise continuous piecewise continuous piecewise continuous and exponential order function for a piecewise continuous and exponential order that means it's going to grow slower than and i like uh an exponential function that's exponential order functions or at least at the level of at most at the level of
7:23:57
exponential function linear exponential function so that's particular exponential order it doesn't grow faster than that so um piecewise continuous and exponential order function exponential order okay so that means it is slower than e to a k t function slower than that maximum is like that ah exponential order function
7:24:27
rf the laplace transform is defined to be the laplace transform
7:24:41
laplace transform is defined
7:24:48
to be uh we can use this notation curly l uh f equal it is from zero to infinity e to the minus s t f t d t so you're getting you integrate with respect to t and the limits going from 0 to infinity we write this one as f s because once you integrate you get the function of s so we write it like that so we use upper case f for that so that is the definition of a laplace transform okay it's just an improper integral you learn in proper integrals in calc too so we use that those idea
7:25:31
okay so let's use the definition to find a laplace transform it's just a simple function so let's say example find the laplace transform of let's say 2 t minus 3 just use the definition to see how it works so the laplace transform of 2 t minus 3 is by the definition 0 to infinity to t minus 3 times e to the minus s t dt you can see this is a good candidate for integration by parts so let's do that so i'm going to use a regular technique you can use a diameter no problem
7:26:13
so if you use a regular technique u has to be 2t minus 3 dv is negative my e to the negative s t d t we take the derivative of this one so d u equal to d t we take the integral of this one is as a parameter is a act as a constant so it is e to the minus s t divided by minus s because you divide by the constant term that's how you integrate a uh exponential function that's a constant you divide by that okay so we have that let's plug in here uh so if you plug in you're going to get
7:26:52
so it's a uv minus vdu so uv so it is 2 t minus 3 that's u v is e to the minus s t over minus s uh in limit going from infinity to zero and then uh so uv minus vdu minus but there's a minus sign there so it's become plus uh there's a 2 i'm going to write the 2 outside so um so vdu so you can see the first product is the top row that's the integral integral is the top row the other one is the bottom row you can see so that's the bottom row we
7:27:32
get now we are writing so is zero to infinity the bottom row so let's try the bottom line in some space so um so let's try the bottom row as the bottom row it is uh so with the minus sign so it becomes plus 2 so we multiply this so i write s outside so you only have negative s t dt that's the bottom row top row is the integral bottom row is the second integral here that's how you remember the integration by parts okay now you can see uh let's so let's plug in here
7:28:04
so if you plug in infinity uh now there are some problems because you see that 1 goes to infinity 1 goes to one goes to zero one goes to infinity because the polynomial goes to infinity e to the minus t equals zero so when so that means in determine uh situation so you have to use the lopitas rule but there is like a small result that you can remember when you try to compare with the exponential function exponential function always win so that means whatever happened the exponential function
7:28:35
everything happened the whole thing the same thing happened that happened to everything so in this case exponential function exponential part goes to zero that means the product has to go to zero so that's a result when you have exponential function and a polynomial exponential function means okay so whatever happens exponential function the same thing happened in the product so that means when you plug in infinity is zero okay so you get zero
7:29:03
uh when you plug in infinity now when you plug in 0 because this is the uh we will apply it for t so you plug in 0 for t uh so you're going to get uh so t term is gone so e to the zeros is gone is one so you have three always but there are two negative sign and also there's a negative sign coming from the when you subtract there's another negative sign that means there are three negative signs so you get negative three over s when you plug in zero okay now the next one ah next one what
7:29:36
you can do uh you need to integrate that so if you integrate what will happen so you have plus 2 of s when you integrate you're going to get one more negative sign so that means it's going to be 1 over negative s e to the negative s t the limit uh infinity and zero the same argument when you plug in infinity it is gone when you plug in zero uh for t so it's one with the negative sign it's become plus so that means uh you end up getting uh 2 over s squared minus 3
7:30:10
over s so that's what you get when you finish the integral and you can see that's not that difficult so this is just use the definition to get this one okay so what's going to happen we are going to do that like every time when you get this one we use the table so let's look at the common few common values uh that you want to remember and in the exam you will get the table okay so you get the uh get a table that you used before so uh these are the common ones uh so table of
7:30:38
laplace transforms this you need to remember it can help you table of laplace transforms so um so let's write the function and the laplace transform so start with one laplacian transform 1 is 1 over s uh that's very easy to find using the definition laplace transform t is 1 over s square those those you you kind of need to remember and then we have t to the nth power which is a factorial n factorial n divided by s to the n plus 1 and then we have sine omega t omega is a constant
7:31:29
uh which is omega divided by s squared plus omega squared and then we have cosine omega t which is s
7:31:46
s divided by s squared plus omega squared and then we have the eat function e to the minus eight t so if you have e to the minus eighty uh it is one over s plus a see this the sign change uh and then we have the uh um heavy side function ut minus a uh which has uh e to the minus a s over s actually you have to be careful um we dividing the whole thing by s okay so it is e minus is divide the whole thing by it's not just the minus is part and then we have the uh
7:32:35
first shifting theorem it is e minus a t e f t so if you get e f t the answer is you just get the laplace transform of the function you replace uh s by s plus a so again opposite sign okay opposite sign so it's a good thing to remember but in the heavy side 5 is the same sign and then the second shifting theorem which is u t minus a f t minus a which is simply e minus a s f s so this is what we call the first shifting theorem and this is what we call the second shift
7:33:27
so these are useful uh when you try to uh find the laplace transform of the inverse laplace so this is the kind of thing that you cannot remember because it's not that difficult if you remember them it's easy to find those inverse and laplace transforms okay so um and then there are some facts about this linear transformation of the laplace transforms one of the big factors is linear this is a linear operator okay this is a linear operator what does it mean uh if you have a sum like that
7:34:06
if you try to find the laplace from the sum it is simply the first one plus or minus it doesn't matter whatever uh the same sign goes here and then g so that is the first condition the second condition is if there is a constant you can simply pull the constant out so those two properties can make life easier uh because it just act like we actually expected so let's try to kind of do our two problems from that so let's do this one uh find the laplace transform of find l f 4 uh the first one
7:34:49
is let's say if t equal 2 t minus e to the negative 2 t plus sine 3 t so let's study for the laplace transform uh soil f is i say 2 times laplacian t is 1 over s squared minus this is the eat function here a equal 2 so that means 1 over is plus 2 and this is sine we know this table is omega over s plus s squared plus omega square so it's a 3 over s squared plus omega squared means 9. ah so that's the answer i mean you you can't do much with this so we leave it
7:35:37
like that uh let's do another one uh how about this one let's say you have f t equal e to the negative 2 t sine 3 t so if you're something like that that's the eat function so according to that your answer is what we can do we first find the laplace transform of the major function the sign 3t so we can do that first it is 3 over s squared plus 9 and then what we're going to do we're going to shift s to s because we have uh uh that's the number so that means the plus two okay so when you have a
7:36:27
remember the format there's a negative sign here but we're gonna go with the plus sign so that means uh it's gonna be you simply replace s with s plus two so if you do that you end up getting 3 over s plus 2 you place s with s plus 2 squared plus 9. if you want you can simplify the bottom that's not a must if you do that if you experiment at 1 at 9 you get s squared plus 4 is uh plus 13. so that's what you get uh so that's laplace transform of
7:37:01
that okay so it's not that difficult but this is how it works okay now let's talk about the inverse laplace transform so we're gonna define the inverse laplace transform kind of in a like a interesting way uh so inverse laplace transform
7:37:26
uh so how we define inverse laplace transform uh we're going to say if the laplace transform of a function is f s if and only if let's go backward ah and forward if you take the inverse of fs is the ft so that's the kind of idea so you can transform backwards so i mean that's how we defined inverse transform that means if you know the laplace transform you can simply find the inverse laplace transform so let's start let's do like two examples just to see how it works
7:38:10
so find the inverse laplace transform
7:38:19
i'll start with this one uh so the first one inverse laplace transform of 3 over s plus 2 minus 1 over s minus 1 square how we do this it's already factored uh otherwise we have to find positive fractions but it's already b given a linearity so it is you're going to take 3 out so it's l inverse 1 over s plus 2 and here you can take a minus sign out it is l inverse 1 over s minus 1 square okay now let's see how we can do it so the first one is easy it is 1 over s
7:39:03
plus a b that's one of the formula in the table so that means this is simply 3 e to the minus 2 t over this sign okay let's say it's 1 0 is plus a minus how about the second one when you want to do the second one you first need to figure out it's a shifting version that's a shifted version you want to figure out what is the base function here the base function is 1 over s squared if the base function 1 over s squared we know that 1 squared is corresponding to
7:39:35
t so we write t because that's 1 0 squared because that's the base function here the base function is 1 over s square you see and then and then you shift it then the shifting part count that means the e term comes so in this case the shifting part e but it is minus 1 so that means it is plus 1 so the plus 1 t so that's what you get okay so it's simple so you need to figure out what is the base function the base function 1 squared 1 over square is corresponding
7:40:06
to t and then you do the shifting that means you multiply by e to the t so that's the eat function example okay so what i use here is i'm going to write here so what i use here is we know that if you are e to the minus a t f t it is simply f s plus a so that's the one we use good um so let's do one more example uh how about this one uh let's see l inverse but three plus two s over s squared plus four so if you get something like that
7:40:49
what we can do ah it's not directly something that we know so what we can do we can split this we can split this into two fractions so what are the two fractions you can see that i'm going to take three out l inverse first one i'm gonna write s squared so what is missing here we know that this is very close to the sign but if you sign the square root of 4 is missing that means 2 is missing so what we do we multiply and divide by 2. so that's what we do we're going to
7:41:25
multiply and divide so we're going to multiply here by 2 and then divide here by 2. so that's the first one we did plus because the linear so things are easy you take the 2 out so we write l inverse s over s squared plus four so now we know all the both so the first one is simply uh three half sine two t second one is two cosine 2 t so it's very simple okay so but we have to uh break it into split and then you can see it's right there it's simply coming from
7:42:10
the table the next major method is the uh partial fraction method so let's discuss that now
7:42:22
so the positive fractions method partial fractions method
7:42:36
so if you don't know much about that there's a video uh that i did for my calc 2 class they're like i explained like like a vast uh majority of questions like in different types so you can look at that and there are like shortcuts a lot of shortcuts how to find the fact uh the coefficients and i discussed that there so maybe you you can look at that if you're enough time okay uh yeah that's the same one i posted earlier also so let's do a problem like this so let's say uh
7:43:06
find the inverse laplace transform of to find the inverse laplace transform inverse laplace transform so like let's say you get like a something like this uh l inverse elements uh 2 s plus 3 over s plus ah i know yeah actually i i'm going to let's let's slightly change the problem to this one because this can be like little bit difficult that's we can do as a second example let's do this one first s plus 2 over let's say s plus 3 um
7:44:01
and s minus 1 something like that but this is easy to do like if you are linear factors those are very easy to do so let's look at how it going to work um so we can write the partial fraction decomposition so it's plus 2 over s plus 3. s minus 1 this identical 2 we can write the linear factors it's a a over s minus 1 there's no order just pick one s plus 3. and then what we can do we're going to multiply the common denominator so we can write this one as
7:44:36
s plus 2 identical to a times the other one plus b times the other one you see you get the other denominator now what you can do just substitute okay that's the easiest way to do this so you can substitute so you're going to substitute let's say you want to find a so you try to make these terms here so that means you plug in s equal one so you plus equal one if you do that you get three equal a uh one plus four this is that a equal three fourth uh this is s plus three so uh three fourth
7:45:19
and then um again uh we're gonna substitute uh s equal negative 3 to find the other one so negative 3 you're going to get negative 1 on this side equals uh the first term is 0 now and then from the other term b you're going to get a negative 4 because negative 3 minus a negative 4. so this says that b equal or 1 4. so that's it so you can find the two fractions are two coefficients so that means if you go back and substitute a equal 3 4 s minus 1 b is 1 4 s plus 3. good
7:46:07
so we have those two now what we can do we can find the inverse now so we can start with inverse which is ah over s plus 2 s plus 3 s minus 1 which is equal by the linearity what we can do we can write this as a two separate one sorry three fourth ah l inverse one over s minus one that's that's one the other one is 1 4 l inverse uh 1 over s plus 3 so now what's going to happen uh s minus 1 we know the inverse of s minus 1 that is e to the t so it is 3 4
7:46:55
e to the t plus 1 4 e to the minus 3 t so that's it so you can see it's very easy to find only thing that we need is partial fraction decomposition so if you know how to do that everything looks great okay let's do one more little difficult example uh so p let's do the other one that we suppose we're trying to do so let's say we have 2 s plus 3 over s plus 1 squared s squared plus 4
7:47:37
so we know how to write the partial fraction decomposition for that so this is we can start with 2s plus 3 over s plus 1 squared and then s squared plus 4 this is identical to so when you have a repeated factor linear factor what we're going to do we're going to start with the smallest one and then the next largest so that's how we do it and then plus this is s squared plus 4 and we know that b has the exact same format as a
7:48:17
when they repeat but here that's not the case uh this is the irreducible factor so we have write c s plus d so sorry that's how we write it uh now what we can do uh the same thing so we write 2 s plus 3 identical to write the missing term here the missing term of 1 of s plus 1 is missing s squared plus 4 is missing here b only s squared plus 4 is missing here c s plus d uh you can see s plus 1 squared is missing uh now what we're going to do we're going to always we
7:48:57
do the substitution first and then the coefficient that's the easy way to do it so we know with the substitution so we can see that uh this term is common so we're going to substitute a c equal negative 1 to find b so we substitute s equal negative 1 if you do that you get 1 equal 0 plus b if you plug in negative 1 square that means 1 plus four uh so you're gonna get so you can see you're gonna get b equal and then also this term is zero so this says that b equal one fifth
7:49:43
so we already find the b value now what we can do you can see that you can plug in anything else and directly find either any other values any other coefficients so what we're gonna do we're gonna go with the coefficients so we always start with the largest coefficient that we discussed several times what's the largest coefficient what's the largest term that you're going to get with s you can see s cube is the largest so we're going to go with that
7:50:08
there's my skip term here uh so that means it is zero uh and then you can see you get a x cubed term and you multiply these so that means the coefficient is simply a the same argument uh from the next one you're gonna you are not gonna get one but from the last one you will again get plus s squared s so that means there's a uh another s cubed term that means c so those are the coefficients that means a plus c is zero that means if you know one you can find the other one
7:50:37
ah now what we can do uh we can write a equation for a squared so what we need is just three more equations because we have three unknown we need three more equations uh i'm gonna go with uh because you nee you know that all of them related with uh uh a so because of that because a has like uh is constant has a is a square has say is cube has a so i'm going to go with not s squared i'm going to go with s okay coefficient of s if you go equation s it is 2 on this side
7:51:15
uh on the other side you're going to get the from the first term the coefficient of s is 4a from the first one and then from the second one there's no uh term so this is easy from the last one let's expand uh this one so if your span is plus 1 squared it is s squared plus 2 s plus 1. so it's going to help you to figure out what is the coefficient so what is the coefficient of s together ah one way to do good is ah you can see uh you can multiply one with c s so it's a c is a
7:51:54
coefficient and to get another s you can multiply d with two s so that means 2d is a coefficient now what we can do we know that a plus c is 0 so that means you can remove 1 of a plus c from this so if you remember 1 of a plus c you only have 3 a plus 2 2d so that's easier compared to the original one and then let's go with the constant term so if you've got the constant term you get 3 on this side where the constant here 4 a is a constant
7:52:28
4b is a constant and from the last one you can see d is a const ah so now what we can do is uh so we need two equations with the same variable so to get that what we can do is you can substitute for b in the last one so if you substitute b in the last one uh what are you going to get uh so you can see
7:52:54
uh if you substitute a b for the last one you can see it's a b is one fifth so it is a four a so this says that a four a it says that for a plus d for a plus d uh equal 3 minus 4b but what is 4b which is 3 minus 4 fifth that means fifteen of eleven so that means you're gonna get equation uh for a plus d equal uh 11 over five now what we can do is uh we can solve uh the last two equations 4a plus 2d and 4a plus d um and if you solve those
7:53:47
last two equations and you can see that you can get a easily because you can multiply the second one by 2 and then subtract the top one so if you uh do that so okay again i just write the two equations here so we have 2 equal 4 a plus 2 d and here we have 11 15 a fifth equals 3 not 4 3 3 and then we have 4 a plus d so what i'm going to do we can solve these two i'm going to multiply the second one by 2 and then you're gonna subtract the first one so once you do that
7:54:36
uh you're gonna get uh 12 over fifth equal is 80 minus 35 a so this is going to give you the a so a equal uh 22 uh a equal actually what do you get 12 12 over 25 so a equal 12 over 2 and a 5 that's what you get for a uh and then we know c because of that then c is simply minus a so this is negative 12 over 25 so and then we can find d from one of those equations uh and then uh we can find d so i'm going to write the value so you can verify so this implies that d equal just plug
7:55:30
into one of those equations uh so it is d equal 1115 over 4a so 11 15 minus 4a so this can give you the d value ah which is 725 okay so we find a b c d also okay that's the only part like taking time uh otherwise just a method is easy so after we know all four so then we can write this expression so we have i'm going to rewrite this one 2s plus 3 over s plus 1 squared s squared plus four
7:56:16
s squared plus four uh equal we had 22 over uh
7:56:25
is 22 or equal actually 125 so it is 12 over 25
7:56:39
one 1 s plus 1 and then we had b is 1 5 1 over s plus 1 squared and then we had 11 a negative 2 12 over 25 c that's the c value and the d was 7 over 25 over s squared plus 4 but you can split these two to get uh not the c s to get a uh sine and cosine term okay now we have everything we need so the inverse of this 2s plus 3 over s plus 1 squared s squared plus 4 equal so so we're going to find the inverse of each term so that means this is going to give you 12 over 25
7:57:46
ah l inverse i'm just writing this one this is not really necessary but for completeness i'm writing and then one fifth l inverse one over s plus one whole thing squared the next one what i'm gonna split this into two so we have negative 12 over 25 l inverse the first one which is s over s squared plus 4 which is a cosine plus 7 over 2 5th l inverse uh 1 over s squared plus 4 so what we can do we can divide by multiply by 2 and divide by 2 that we can do so that means your final answer
7:58:30
is simply going to be 12 over 25 e to the negative t plus a one fifth t e to the negative t that's the second one how it coming from because s plus 1 that means 1 0 square s plus 1 square that means 1 over s squared is the term that means t and the shifting is e to the negative t that's a shifting term and then we have uh negative 12 of 25 the cosine 2t and then here what's going to happen is the sign 2 is missing so you multiply and divide by that so that means
7:59:11
you're going to get 7 over 50 sign to t okay but uh just like this may be like little long uh to an exam so you will get something uh you can manage um this is like little long compared to other problems okay good but i want to show you like how it works it's not that difficult only problem is the how to find the partial fraction decomposition um good that's about that and then let's talk about how to solve a uh differential equation using the laplace
7:59:48
transform method that's the most important one so solving uh solving differential equations using laplace transforms a method
8:00:09
solving differential equations using the laplace transforms method uh so let's start with the laplace transform of derivatives laplace transform of derivatives ah we know these results so the laplace transform of the first derivative is s f s minus f 0 laplace transform of the second derivative is s squared if s minus s f 0 minus f prime you can see first term is positive everything else is negative this is something you can have so if you write the third derivative so
8:01:01
you can just keep doing it if you write the third derivative uh which is s cube f s minus the power of s goes down so that means squared is zero and then s f prime 0 minus s double prime 0 so you can see all the terms after the first terms are negative so that's something you can remember okay these terms are negative okay and then so you can continue like that so you can have the general one so it is a laplace transform of f to the nth that is so the nth derivative is s n
8:01:44
f s minus s to the n minus 1 power f 0 minus s to the n minus 2 power f prime 0 like that and then you're going to give me uh to the n minus first derivative zero okay so that's the idea so you can see that except the first term everything else is negative so there's a negative sign okay all negative only the first one is positive good uh so now let's see how we can use this uh to solve a differential equation so let's do also solving differential equations so let's look at this example
8:02:37
so uh solve
8:02:43
x double prime minus x prime minus 6 x equal let's say 5 e to the negative 2 t along with x 0 equal 1 and x prime 0 equal negative 2 so if you get something like that so what we normally do is uh we can take the laplace transform of the equation that's what we're trying to do and then we can use the initial conditions given um so so the answer so let x s equals laplace transform of the expected solution and then what we're going to do our table laplace from the equation so taking
8:03:36
uh laplace transform of the equation you're going to get s squared x s
8:03:48
minus s x 0 this i small minus x prime that's the first term the second second term says minus there's a minus sign so let's plug in here so s x x minus x is a good idea to write this one uh to avoid uh errors and then minus uh six excess equal laplace transform of that term is a five and this eight function so it is s plus 2 so that's what you get now what we're going to do we're going to substitute the conditions that we know we know that
8:04:28
this is uh this equals to 1 is equals to 1 and this equals to a negative 2. so if you plug in that ah then you can bring so you have a negative x term you can do the other side and we have negative 2 so it becomes positive 2 and positive 1 is 3 so we have negative s plus 3 so you can bring it to the other side take the command uh so we have a squared here and then we have minus s and then we have we have negative 6.
8:05:02
uh excess and you can see this is exactly equal to the characteristic equation but with the s not the lambda but the s so that that happened always and then now we're gonna bring the other terms to the side so we have s minus 3 plus 5 over s plus 2 and now what you can do we can divide by that and you can see this we can actually factor what are the factors of this uh the factors are s minus 3 and s plus 2 those are the factors
8:05:41
so let's divide so when you divide what's going to happen we have s minus 3 here and then we have s minus 3 ah s plus 2 plus 5 over so we have uh s minus 3 yeah is plus 2 and there's another x plus 2 already over there so that means this can be s plus 2 whole thing is squared and the interesting thing is sometimes uh see this could cancel so it simplifies well so we only need to find one partial fraction but otherwise what we're going to do we're going to simplify this here
8:06:26
so we normally what we do here here is we simplify this we simplify this before we find the partial fraction so that it's going to make the calculation much easier because it's going to be an internal calculation that calculation gonna like you know take care of most of the terms so it's always a good idea to simplify this first but here you can see um we don't need that because that term cancels so we have like nice one already uh so we
8:06:52
can separate those two terms okay so uh so what we have is 1 over s plus 2 so we leave that there because we already know the inverse of that and we only need to find the partial fraction of this term s minus 3 s plus two whole thing square normally what we do we're going to simplify this because it's going to make the calculation easy so let's try to find the partial fraction of that uh so now we're gonna go with the partial fraction of that 35 over s minus 3
8:07:26
s plus 2 whole thing squared which is equal to a over s minus 3 plus b over s plus 2 plus c over s plus 2 squared uh
8:07:47
so we're gonna uh so and then after that we're going to get 5 equal a the missing term missing term is s plus two whole thing squared and the next one the missing term is uh one of the s plus two is missing s minus 3 is missing last one only s minus 3 is missing so we can use a substitution technique now always use substitution first so we can substitute let's say s equal 3 if you do that you get 5 equal you can see plug in 3 this goes away so you get a so a if you plug in 3
8:08:28
this is s equal 3 so it is 3 plus 2 squared plus 0 plus 0 this is going to give you a equal 5 over 25 so 1 5 that's a now we're going to substitute uh let's try to find uh find what is other common term you can see this is common see so that means you can find c so applying a sequence negative two so you apply a c equal negative two uh five equals zero another zero and c if you plug in negative 2 this is negative 2 negative 3 this can give you uh 5 divided by 5 the c equal negative 1
8:09:10
okay and then ah so after so we only need to find one more what we normally do is find the equate the coefficients of the largest term remember that okay because that's going to make the calculator much fast so coefficient of the largest term what's the largest term squared so we're going to go with the coefficient of a square the largest term so it is five largest uh degree uh so you can see how to get a square term the first one
8:09:39
going a square term the coefficient is a in the next one when you foil you get a square b so that's the next one since we only know a we can find b from this one so this says that b equal five minus a ah so that means ah
8:10:03
ah no actually so what i did there's no zero on here so there's a problem here uh so you can see there's no uh there is no a squared term so this is zero on this side okay there's no square terms it's zero so you get b equal minus c so since you only know a so it is negative one fifth sorry for that okay this is zero there's no square term on this side so it's zero so so you could so you could a b three like really quickly now what we can do we're gonna substitute that uh so that means
8:10:36
uh so is one over 5 phi over
8:10:46
5 over s minus 3 s plus 2 squared equal you cannot equal now it is one fifth uh s minus three
8:11:05
uh actually let's write the ss term it's not this one because we already know uh yeah so let's write s1 it's very important so if you have s term so what do you get uh so x s uh which is 1 over s plus 2 plus the partial fraction of these okay so we're going to have the partial fraction now this is 1 0 s plus 2 we had already so let's try the partial fraction here so it is 1 5 uh over uh one fifth or s minus one fifth two is minus three so yes minus three that's a and then we have b
8:11:47
b is negative one fifth over s plus two and the c uh which is uh negative one so it is negative one over s plus two squared that's what you get and you can see that you can combine these two so they're going to simplify so they're going to simplify so you're going to get uh
8:12:16
so you're going to get four-fifths so we have four fifth s plus two uh plus one fifth s minus three minus 1 over s plus 2 squared so we can find the x t now so x t is simply so i just wrote like completely but some steps you can just ignore okay i'm not serious so this is the inverse of this so we're gonna go term by term so the first one says it is uh 1 4 5 e to the negative 2 t next one says one fifth e to the three t the next one is you can see it is the
8:13:04
the base function is one o square that means the t e to the shifting is negative two t so that is the solution of the differential equation but this one is not that long uh it like kind of came uh quickly uh even like it's not that difficult to find the pulse or fraction decomposition for that so this is a kind of good problem uh especially for a final exam good uh next we can talk about the convolution method
8:13:38
so let's discuss the convolution method convolution method
8:13:53
so we know that the laplace transform is linear that means if we have a sum uh we can write this one as lf plus lg so we can do that but if you have a product f times g this does not equal to the laplace transform of f times laplace transform of g so there's no such result so instead we have what we call the convolution so let's say let's look at the definition so what we call the convolution product convolution product
8:14:44
so the convolution is defined as an integral so the notation is star so f star g that's the convolution of f and g so which is from zero to t what we gonna do we let's introduce a new variable called tau so it's f tau g t minus tau so t is the variable sitting there and d tau oh you can even switch the order so you can see you can put the t minus tau for the first one and then g tau for the second one the interesting part is they are you get the exact same answer from them
8:15:22
so in other words this means that uh this product is uh so you can you can switch the two so that means it is uh communicative so you can write it like that that's what means you can switch the order you still get the same answer so now the question is uh we put t minus tau for the easy one so that is easy to integrate so that's how you're going to pick it so whatever the easy one so we're going to see when you look at a problem it will be
8:15:51
clear what we mean by easy one and then we have this major result so the theorem is that she says let's say f g ah piecewise continuous uh piecewise continuous piecewise uh continuous and [Music] exponential order
8:16:15
ah and the exponential order and of exponential order
8:16:24
we need that to guarantee that the laplace transform exists then what happened the laplace transform of the convolution product f star g is simply the product of the laplace transform of f and g so in other words we write this one as f s times g this is just the product so this is the major result uh this is very useful uh so that's what we have so if you look at the convolution laplace transform of the convolution you're going to get the product of the laplace transform
8:17:04
so the benefit is this is very useful when you want to find the inverse so that means uh so let's write this as a not so that means so if you have inverse of the of a product of two transforms if you know that the two functions what are the two functions are just take the convolution so that's the idea so this might be much faster compared to uh partial fraction or other techniques so that's that's how we can use it so useful are useful when finding inverse useful in finding a inverse
8:17:46
inverse laplace transform so let's uh look at that in a in a problem so let's try to do this too so example i find the inverse laplace transform find the inverse laplace transform so let's do the first one let's say l inverse 1 over s plus 2 times s squared something like that so we know how to do this one using password fraction method we write the form find the uh abc i believe there are three terms now and then invert but how
8:18:37
to use the convolution so we know that so we're going to kind of try to identify so we can write this one as 1 over s plus 2 times s squared so we isolate that now we can see ah this function is corresponds to e to the minus 2t this one corresponds to t so those are the two functions uh and then if you tell a person those uh that's what you get so that means according to the theorem this is simply the convolution between those
8:19:12
two functions so that's the idea just take the convolution of those two so now what we can do we can use the definition and try to find it so the question is which one uh we're gonna put t minus two so when you have exponential function it's always a nice idea to put the t minus two to the exponential function so so we can do that so it's 0 to t e to the negative t we put t minus tau here for the exponential function other one has just become tau d tau so the reason is you can see you can
8:19:45
pull out e to the negative uh 2t because that's a constant now uh and then 0 to t uh e to the negative not negative you become negative positive because you can see there's a negative negative so e to the 2t tau theta so this is a very easy integral using partial integrating by parts so let's do that so it is e to the negative 2 t now you can just integrate my parts so we're going to we can have the di method for this one so let's use the di method d and i so we're going to say for the d
8:20:20
v with tau i we put e to the 2 uh tau so take the derivative 1 take the integral one half e to the two tau take the derivative zero so that means one fourth e to the two tau and then what we do we multiply plus minus so that's what we get that means the answer is simply uh we're going to multiply the two so it's a one-half tau uh e to the two tau and then minus uh you get 1 4
8:20:59
e to the 2 tau and then you can evaluate that let's change that we can evaluate that at t and 0. so that's it so that's the answer uh so what we get so you're going to get e to the negative 2 t when you plug in t so you get
8:21:28
you get one half t e to the 2t and then minus 1 4 e to the 2t and then when you plug in 0 the first one is gone the next one becomes simply one-fourth and then what we do we multiply by e to the negative 2t so you get one half t uh minus no plus uh yeah so it's gonna be plus one fourth ah
8:22:10
good one fourth and then uh we have no that's that's uh actually so it is negative because we we multiply negative 2 t with 2t so you get negative 1 4 and then the last term going to be plus 1 4 e to the negative 2 t so that's the answer using the convolution method so it looks like it's little faster compared to the partial fraction because we just need to evaluate a simple integral here good uh good so let's try to do one more so let's do this one so the laplace inverse laplace transform
8:22:49
of 3 over s squared s squared plus 4 so this one if you do the regular way it little lengthy but let's see how uh it's going to be now uh so what we can do is we know that uh if you look at square plus four that's related to the sign so what we do we take uh two uh we we multiply and divide by two uh so this is l inverse so we're going to arrange it so 1 0 squared times 2 over s squared plus 4. so we can write it like that now you can see this is t this is related to
8:23:33
sine 2t so we have the two functions now what we need is according to the theory this simply becomes three halves that's there t convoluted with sine 2 t so that's what you get now what we're going to do we need to find the simple convolution and you get the answer so that means uh this is equal to so we have again we can use the uh dm method for that so the d and the i so d uh first let's write the integral representation is a two third
8:24:13
three halves zero to t uh i'm gonna put t minus star for the first one and then sine two tau d tau so this one we can use do using the uh di method so t minus tau goes here sine 2 tau goes the tau is the variable so we still the derivative t is a constant so you get negative 1 here and the other side you get negative one half cosine two tau and then take one more derivative zero so uh so cosine that means you can divide by one half so it's a negative uh one fourth side 2 t 2 tau so plus
8:25:00
minus so we can write that now so you're going to get so let's plug in that so we know the two product here so let's plug in that so it's three halves uh the first one is uh negative one half a t minus tau cosine 2 tau a minus 1 is a because we have 3 minus sin this is minus 1 4 sine 2 tau we can evaluate that at t and zero so if you evaluate if you plug in t the first term is zero
8:25:45
uh and then if you plug in 0 so you get our t half and then from the next one when you plug in um zero you get zero of number t you get negative 1 4 sine 2 t and then finally if you multiply through you get 3 4 t minus 3 8 sine 2 t it looks like simple compared to the partial fraction method uh that little end so like it's up to you like you know whether you use a convolution or partial fraction method we need a problem unless it says that use the convolution so that's the kind of
8:26:40
idea um so how about this one so how about this laplace transform so this is another example so find the laplace transform of the integral from 0 to t sine 3 tau theta so you can see that this is a convolution problem what happened why you don't see the t minus tau term because as a constant term so that means it is simply convoluted with one so that's why you don't see the t minus two because you can put the t minus store to the constant so that means this is simply the convolution between one
8:27:24
and sine three t so if you do that one uh so uh so that's the convolution so actually we will write it like that so this is the convolution of this outline this is the convolution between uh 1 and sine 3 t so by the product this is the convolution of 1 times the convolution of sine 3 t so if you use laplace also 1 is 1 over s times 3 over s squared plus 9. so if you write s just one single expression s times s squared plus 9. so uh so we can do that so like sometimes it's not really
8:28:16
visible you know but it's there yeah so that's a good one and then uh so we can we can use convolutions if you want to find the inverse okay so let's uh so let's put it off in word so we can say uh use convolution
8:28:38
not a must but you can okay use convolution when finding inverse finding inverse laplace transform so let's go to the next topic that's the heavy side function
8:28:59
a heavy side function and switching
8:29:10
so we talk about the switching you know when you have a piecewise function you can use this as a switch so let's go with the definition of a heavy side function
8:29:24
oh you need step function that's the other day so if you got a graph it looks like that so what's going to happen until 0 is 0 and then at 0 it can jump to one and it stays at one so that is uh uh the what we normally write as ut at zero and then we have the shifted version of that so let's say it's gonna jump at a so until late is zero and then at one at a is going to jump and stays there so that is the uh heavy side function shifted by a sorry ut minus
8:30:02
c um this is one at that point so that's the function good now uh so the laplace transform of those so we have the laplace transform of the huge side function so laplace ah transform so we have laplace transform of u t minus a equal e to the minus a s over s so that's one it's very useful um it's very easy to show or this one just use the definition and then we have also second shifting theorem second
8:30:50
shifting theorem so what it says that the laplace transform of u t minus a times f t minus c so the exact same form at t minus a t minus c so see that t minus a uh t minus a term and t minus c so what's going to happen this simply become e to the minus a s so they have the exact same format again minus s times the laplace transform of s so that's uh that's what we call the second shifting theorem ah now let's try to use this to find the
8:31:29
inverse laplace transform of a certain function so let's do this one just to see how it works for example find the inverse uh laplace transform of this one so l inverse s plus 3 over s squared plus 4 e to the minus 2 s something like that so let's say you have this one so it's clear that that is the exact same format that we have uh the second shifting theorem version so what we're going to we're going to shift this first
8:32:12
so we can write this one as an inverse uh we can go with s first so s over s squared plus 4 e to the minus 2 s and then look at the next one the next one we have 3 but we actually need 2 so we can do that 3 halves l inverse 2 and then we have s squared plus 4 e to the minus 2 s like that now we can see uh this is the cosine ah 2t this is the sine 2t so what we will do is uh from that we need to just shift this so that means the answer is you can shift by 2 so it is u t minus 2 the cosine
8:33:06
2 times t minus uh 2 plus we have that 3 halves again u t minus 2 sine 2 t minus 2 so that is the laplace transform so that is how we use that idea when you have a e to a function like that it's just a shifting you just uh use a second shift in there for that now uh let's look at how we use this for uh switching um so let's do a problem like this uh let's do uh rewriting a piecewise function using uh switches so rewriting
8:33:58
rewriting a piecewise function piecewise function regarding p systems okay ah so let's look at the idea like a problem like this so example uh uh so let's say you have a function like this so let's say rewrite okay are using a heavy side function
8:34:40
so what we can do let's say f t let's see how piecewise function like this t minus 1 on the interval 0 to 1 and then sine t after that so if you look at the graph this is how it looks like so from 0 to 1 it is a t minus 1 so that's a t minus 1 function this can go with 1 negative 1 and then what happened ah it's gonna jump to the sine function at one so sine one sine one is a number somewhere and then it's gonna follow the rest of the sine function so it's gonna go like that so that's the
8:35:29
idea uh after 1 it's going to follow the sine function but until then it's t minus 1 so this is these are what we call piecewise function so this is a nice piecewise function so the question is how to write this one as a using uh as a single function as a single function using um a heavy side function so what we can do we can add a switch here so you can see v1 into the switch for the first interval from zero to one the is switch on at zero and then off at one
8:36:03
so that's what we do and the next one gonna switch on at one so that's the idea so that means we can write this one as we can multiply just by the switch so we have the function t minus 1 we're going to multiply by the switch so the switch here is u t minus u t minus 1 and then plus that's the first one that's gonna switch on only in that interval the next one is sine t uh thus you're gonna switch on after one that means you multiply by t minus
8:36:37
one the function whatever the function so sine t so that's how you write are using switches and if you want you can even combine them into one function for example there's only one ut function so that means you can write this one as t minus 1 times u t and then there are t minus 1 function there's a t minus 1 here and then also sine t that's positive so what we're going to do we can write this one as plus 1 minus t plus sine t times u t minus 1 as two separate
8:37:14
u functions so that's how you write this using switches simple idea you can switch on at the first node off at the other node so that's the off part that's the odd part uh yeah just pick the two values so that's the idea and then after what you can do you can use this theorem to uh proceed with this so let's just solve one example to see how it works uh probably we can combine this and the convolution in this problem so let's solve this one uh so solving
8:37:52
solving a differential equation solving a differential equation
8:38:01
equation with a piecewise input
8:38:10
piecewise input
8:38:15
solving a differential equation with the piecewise input like that so the example is this so let's solve this one solve let's say x double prime minus x prime minus 2x equal f t along with x0 equals 0 and x prime 0 equals 0 just to make things easy so we are let's say if t equals something like this 3 t 0 t greater than equal 1 and 0 otherwise so that means if you look at the graph this is how it looks like so what happened until 1 is 0 at 1 it's exactly start with the
8:39:16
3t graph so that's that it's going to follow that so that's the if how the st function look like at one jump to that and until there is zero just stays zero so no input and all of a sudden it's going to exert a uh some input like voltage current something like that every changes uh linearly so now what we do if you get a problem like that we first use the switches to write it as you can see you want to switch on at 1 so that means this is simply 3 t u t minus 1 that's what 3 t times
8:39:55
u t minus this is this is switch on at one so the goal is to use the second shifting theorem because you're gonna try to use the because there's no other way to do it ah you can directly use the theorem like a integral that you can do you can use the integration and you can integrate this one that's one way to do it but if you use a second shift in theorem you have to write this in t minus a form t minus a form so a in this case is 1 so let's do that so what we're gonna do
8:40:31
we're gonna subtract and add one so we write this one like this trick here so what we're gonna do we'll start with t we're gonna subtract one and add one and then u t minus one now if you distribute with the first one you get t minus 1 u t minus 1 in this case the function is f t that's the function if t equal t and then the other function is simply 3 times u t minus 1 we know the laplace transform of that so that's how we find so we're gonna
8:41:10
simply get the two laplace transform of these two uh if you write it like that so that's the that's the point uh we write it in this form okay ah so so the answer here
8:41:30
because you're writing f t in this form so we write it like that now we can simply find the ah the uh laplace transform so so if this is the case then the laplace transform of f t is uh so it's a 3 times the laplace transform of t minus 1 u t minus 1 plus 3 times the laplace transform of u t minus 1. and we know there are formulas for both so the first one is simply 3 e to the minus s over the function is 1 0 squared that's the laplace transform of 1 y
8:42:26
1 because that's the function plus the second one it is 3 times e to the minus s that's the second one so what we can do now over uh s actually so there's s so now what we can do take the common denominator common dnd is squared so we can write this one as 3 s plus 3 over s squared e to the minus s so you can write it like that okay so that's the laplace transform of the signal the input function now let's take the laplace transforms the differential equation so ah so we have are the
8:43:12
inverse not the inverse the laplace transform ah no yeah so let's let's do it like that so let's say uh taking the laplace transform taking the laplace transform
8:43:36
uh of the differential equation we're going to get a squared sx there are no initial conditions so only the s sx terms minus and then minus 2 sx that's what we have on the other side is the ss that we already know so this is s squared minus s minus 2 times s x the other side it is 3 s plus 1 over s squared e to the minus s ah so what are the factors of this s minus 2 and s plus 1 so that means when you divide s plus 1 get cancelled so so you simply get sx s plus 1 get cancels you get 3 over
8:44:33
s squared s minus 2 e to the minus s so if you have e to the minus x what we normally do is first try to find the laplace transform of the rest of the function and then do that then use the shift second 15 theta our transform so that's what we normally do so to find the ah inverse laplace transform let's use the convolution now ah so we need let's say we need the inverse laplace transform of 3 over s squared s minus 2 uh so this is a linux we're gonna
8:45:20
break this 3 over square times 1 over s minus 2. so this simply says that it is 3t convoluted with e to the 2t so what you get now we can use the definition so 3 times uh when you have exponential function that's where you're going to put the t so it is 0 t minus tau so it is tau e to the 2 t minus tau d tau so we did this a lot so now we know what to do with this so it is
8:46:09
3 e 2 t we take it out 0 to t tau e negative two tau d eta that's what you get ah and then we can integrate this one using the d i method again so so we have tau e to the negative two tau so one zero uh one half e to the negative two tau and the mean one more so it is one fourth e to the negative two tau you multiply these two and multiply those two plus and a minus so now we can write the answer here so this is simply going to be 3 e to the 2 t times let's write the first product so it is
8:47:02
one half tau e to the negative 2 tau next one is one half a negative one fourth e to the negative two tau and then you're gonna go with t and zero
8:47:20
so if you plug in the two so what happened is so you have 3 e 2 t let's plug in t so you get 1 half t e negative 2 t and then when you plug in 0 is gone and from the other one when you plug in t you get 1 4 e to the negative 2 t and again 0 is positive 1 4. so that's what you get and multiply through so you're going to get you multiply by that so 2 t get cancelled uh because the negative sign so you get the first one is negative 3 halves t next one is 3 4 and the next one is plus 3 4
8:48:18
e to the 2t so that is what you get as the uh inverse laplace transform not now what you want to do is you can use the second shifting theorem uh to shift okay so by the second shift in theorem so by
8:48:38
the second
8:48:42
shifting theorem now we have x t which is the inverse of 3 over
8:48:59
s squared s minus 2 e to the negative
8:49:07
e to the negative s so that means you can shift by negative 1 so that means everything would be so so we get the function first so this is the function so we're going to go with the function you replace t with t minus 1 negative 3 for there are no t here and then 3 4 we have e 2 t minus 1 and then everything multiplied by the heavy side function u t minus 1 so that is the final answer for that problem yeah to get the same answer you can also use the uh password fraction with the test field
8:49:52
okay so you may use you may also use the partial fraction method
8:50:11
good so now let's try to go to the last topic of laplace transform that's what you call the direct delta function so let's talk about that uh so the direct delta function impeller response so the dirac direct delta function
8:50:32
direct data function and impulse response impulse response
8:50:47
ah so let's talk about what's rejected the function or this impulse so like you know the sudden force acting like a at a single point that's the kind of idea so uh we can write this one as delta t so like a
8:51:07
so uh is acting at the zero points is called delta t and then ah we have the shifted version of that so the shifted version is uh at a that's impulse at a so that's it and then everywhere else it's here so this is what we normally write as delta t minus is impulse at a and then we have the laplace transform of that so the laplace trans the first one is delta t that is one and then the laplace transform of the shifted version delta t minus a uh which is e minus a s so this is useful
8:51:56
uh to remember and if you plug in a equals zero ah you get the other one good i mean that's it so let's try to see how we can use this uh to find the inverse laplace transform so let's do a problem like this uh find the inverse laplace transform of uh so let's say find
8:52:21
the inverse
8:52:25
plus transform of uh the first one l inverse s plus 3 over s plus 1 and you can see that this is not a proper uh
8:52:50
this is not the proper fraction so if you have proper improper this is improper fraction what you can do is you can do the long division and find the remainder and the quotient that you can do also you can use some kind of tricks here so that's much faster so what we're gonna do we try to create the bottom term in the top that's the this argument so how we do that we can write this one as is we write this one as s plus 1 plus 2 over s plus 1 then we can write with this
8:53:25
and with the other one so we're gonna we're gonna separate this now so we can write this one as
8:53:34
ah so l inverse s plus one over s plus 1 plus 2 over s plus 1 so that means this is simply l inverse 1 plus 2 over s plus 1. we know that the 1 that means a delta function this is delta 1 uh it's a delta t that's 1 plus 2 over s 2 o s plus 1 means e to the e to the negative 1 t because a equal 1 in this case is 2 e to the negative t so the certificate so that's how we find the inverse when you have an improper uh fraction like that so we so their delta function comes into play in this case
8:54:24
so let's just do one uh problem also another problem like that so how about this one l inverse so let's say uh s listen look uh s e to the minus 2 s plus 3 over s plus 1 um so what we can do with this again we can split this so this one we can write as l inverse so you can see we have s here so we can do the same idea so we can write this one as we try to create the denominator so it's plus 1 minus 1 uh e to the minus 2 s plus 3 uh so this is over s plus 1 over s plus
8:55:25
1 you can already like that now the same trick so you can go with s plus 1 with the bottom term so that means we can write this one as uh if you go is plus 1 you simply get e to the minus 2s if you go with minus sign you get minus 1 over s plus 1 e to the minus 2 s and then we have this 3 s plus 1 now we got term by term so the first one is simply delta t minus 2 where a is in this case 2 so is the t minus 2. the next one is ah so we had to use the heavy side function for this one we find it and
8:56:14
then we could transform that you can see you have to the function is is 1 0 s plus 1 s plus 1 that means e to the minus t you replace t by t minus 2 and multiply by the shifting function as well so it is e to the minus t you replace t by t minus 2 so replace t by t minus 2 multiply by u t minus 2 just that and the next one is simply 3 e minus t so that's the kind of idea so this is how we can do it when you have a term like this this is special case so you
8:56:54
uh you either get this or you either get that so it's gonna come with one of those good now let's try to solve uh one problem uh using this method that's the end of uh laplace transform techniques let's just do one example uh so find find the impulse response find the impulse response find the input response of a system
8:57:33
okay so for example let's say so this is the question so i'm going to give the exact same wording find the impulse response or simply say solve the differential equation find the impulse response of this one x double prime minus x prime minus x equal let's say 4 delta t minus 3 something like that okay so where this is coming from and also with x0 equals 0 and x prime 0 equals 0 just to make things easy so where this difference equation coming from so this is a kind of situation
8:58:16
let's say you have a mask on a table which connected to spring with the friction uh stable now what we're gonna do you're gonna give a certain impulse to one direction at the time t equal three so you wait and just give a sudden uh force in uh one direction so that's the kind of situation so now you get a difference equation like that uh so let's try to solve and see what's gonna happen so what we normally do so just the same procedure take the uh
8:58:56
uh laplace transform take the laplace transform so you get s squared rsx plus s x s minus x s equal okay let's write the laplace transform of that we know that just the e minus whatever the number is 3s very simple uh initial condition 0. so you don't get that so here we have s squared uh it's a minus sign yeah this is minus squared minus s minus 1 times sx equal 4 e negative 3 s
8:59:45
so six equal four e minus three s over uh s squared minus x minus one and you know that you can see that there are no factors for the bottom function so what we can do we'll go with the completing square technique uh so let's do that so it is 4 e minus 3s what's the completing square here so it is s half of the s term half of the collision is turned in one half we can square that so it's going to give you the first two terms and we it also give one fourth uh but there's a
9:00:25
you have to get negative one so that means you add subtract 1 4. so if you subtract 1 4 this can be 5 4. so that's what you get now this is you can see this you can invert uh so so what's gonna happen now is uh yeah actually uh i'm gonna change this uh to that number so it's a change just a plus plus plus and a plus and a plus so it's simply changed to the change plus i just make things easy uh i don't want to make it too complicated it's already complicated so now what we can do are we going to
9:01:08
kind of clear the terms so you can see this is one plus one s plus half squared minus so what is this is a sin function this is shin uh we know that there's a form of a hyperbolic sine uh so what is missing is square phi over two so we're gonna uh multiply and by and divide by that so it's a four so we are missing square root five or two so we're gonna divide and multiply uh good so yeah i didn't write it properly so what we're gonna do uh so we can write this one as
9:01:54
s plus one half squared minus five fourth so square root phi over two is missing and then we multiply outside upside down of that that's what we do and then we have also e to the minus 3 ester so this is simply the hyperbolic sign and then we're gonna do the shifting part so that means the final answer so we can see so so this simply gonna this simply corresponds to e to the negative one half t sin uh square root 5 over 2 t so that's the function coming from there
9:02:41
that you know already now what we can do we can use the e to the minus 3s to add a switch here so taking inverse laplace transform taking inverse laplace so what we have is x t equal 2 square root 5 u t minus three u t minus three e to the minus one half t uh so what we're gonna do what's going to happen i have shift by minus 3 so t minus 3 comes there and then sin square root 5 over 2 this one also t minus 3 this is because of the e term this coming from the e term
9:03:35
uh good so that is the answer for uh this problem so you can see that there is suddenly start at uh after three then that's when the impulse so i mean you can see this there's the whole problem so this is like a kind of simple problem compared to other problems okay good so that is the end of laplace transform method next we're gonna do uh we can learn how to solve systems so we can talk about some examples there
9:04:08
in this uh we're gonna uh talk about how to solve system of equations system of oris in this case so we can talk about three methods the first one is the elimination method elimination method so there are two type of problems that we can uh try uh so there are two types we're gonna go one by one and then the next one is the laplace transform method laplace transform a method so this is actually uh kind of sophisticated because you can do
9:04:53
you can handle a lot of cases with that and then the third one is the eigenvector method
9:05:06
eigenvector r method so under the eigenvector method uh we're going to consider two cases we can do two by two uh two by two case is very easy and then we can do three by three when you do three by three there are actually three cases to consider um uh one is there are two eigenvalue case
9:05:33
uh that's kind of easy very easy to find eigenvectors we can use shortcut for that so we will use the shortcut and then there are three eigenvalue case three eigenvalues
9:05:53
uh three eigenvalues
9:05:59
in that case what we normally do we use the eigenmatrices for that i get matrices that's the new concept we discussed that earlier in a different video so you can look at that and then there's a case also uh what we call defective matrices and we are not going to discuss that in this but uh we have that in the in another video so the defective matrix defectory matrix but we will not discuss that here so we gonna discuss one from two by two and one from three by three
9:06:38
uh with two eigenvalues so that's what we can do okay so let's start one by one now let's do the elimination first so the elimination method
9:06:52
elimination method
9:06:58
so there are two types so let's start with type one uh so let's do that with the problem so let's say solve the system solve the system ah so let's see three dx over dt minus d y over dt and on the other side you can see some function of the independent variable so let's say just put something so like let's say sine t and then we have another one dx over dt plus 2 times d y over d t which is equals to 2 e minus t minus 3 let's say cosine 3 t plus 1 so what we can do we're going to
9:07:56
call this equation 1 and equation 2 for references uh so this is like just like how we do this like as like when we solve linear systems like when you have like a with several variables if you see like linear systems we know how to do we can just eliminate [Music] one variable so we can use the same argument for example in this case you can see when you see plus and minus sign that's the kind of we're going to pay attention to because that's going to
9:08:28
make it easy so what we're trying to do is uh so in theory what we do eliminate one of the x prime of i prime so that's the kind of goal so eliminate uh eliminate one of x prime or y prime uh for example uh when you see plus minus sign that's we're going to pay attention to that for example if you multiply the first equation by 2 and add it to the second one you can see the d y term gonna go like eliminated so that's what we're trying to do so we can see by 2
9:09:02
times the equation 1 plus 2 if you do that if you multiply the first equation by 2 and add it to the second equation you can see the y term get cancelled so you're going to get 2 times 3 that means 6 plus 1 so you get 7 dx over dt are equal so you multiply the first one by 2 so that means 2 t squared minus okay let's look at what's going to happen here so we're going to multiply this one by 2 so you get negative 16 and you add there are in the second equation you also have 2 that means you end
9:09:34
up with 14 e to the negative 2 and then we have 2 sine t you're going to get in the bottom you're going to get 3 cosine 3t and then plus 1. so that's what you get just a long expression because there are many terms just integrate now once you integrate you get x so x equal the integral of uh dx over dt uh actually before we before we do that we're gonna divide this yeah let's write that same step so dx over dt if you divide by 7 um you can see it is 7
9:10:11
it is 2 over 7 2 over 7 t squared minus 2 e negative t plus 2 7 sine t minus 3 7 cosine 3 t plus 1 over seven so that's what we get uh now what we're gonna do uh uh we're gonna integrate uh the two sides so just integrate with respect to t you get x okay so now what we're gonna do integrate so once you integrate you get x which is the integral of dx over dt from calc 1 so just integrate so we have a lot of terms so integrate term by term so we have 2 over 7 t squared i'm going to write
9:11:02
everything but you can skip this step it's not really necessary so it's negative 2 e to the negative t plus 2 over 7 sine t minus 3 over 7 cosine 3 t plus 1 over 7 dt so if you integrate i'm going to write the answer here so you're going to get you have to divide by 3 so it is 2 over 21 t cubed minus okay when you have exponential function you divide by the constant so negative 1 so that means can be positive 2 e to the negative t sine means negative cosine so it is negative 2 over 7 cosine t
9:11:51
in the next one this is the cosine but this is a 3 that means you divide by 3 it's a cosine you can sign so it's negative 1 3 1 7 sine 3 t because you divide by 3 plus 1 over 7 t plus c1 introduce a constant name so that is the solution for x now what we can do there are two ways to do uh two ways to find y one way is you can plug in this solution in one of those equations so if you plug in that solution one of those equations you can see that the dx over dt can go away and then
9:12:30
you only have dy over dt then you integrate that to find y the other way is just do the same thing as we did before to uh eliminate the extra dx over d theta and then you're gonna get d y d t and just do the same thing so um so i'm gonna write this find d y over dt as before there are two ways
9:12:59
two ways to do it okay so we can use one of the methods so we're going to buy 1 minus 3 times 2. why we do that then you can because you have a 3 and a 1. so if you multiply the bottom one by negative 3 and add uh dx or dt can go away uh so that's exactly so we're gonna get negative seven we multiply the top one by um bottom one by three so you get negative six and there's another one so it's negative 7 d y over d t and then you're going to get t squared minus 14 e to the negative t plus sine
9:13:47
t minus nine over seven cosine three t minus three that's pretty good uh okay and then what we're going to divide by 7 as before so you get d y over d t equals negative 1 over 7 t squared positive 2 e to the negative t negative 1 over 7 uh sine t and then yeah it's again negative uh negative two nine of uh
9:14:38
uh actually uh yeah so there's a i don't know whether it's negative seven nine or seven it's just nine so you have plus nine yeah uh because we multiply um so let's look at that one again uh so we have yeah so we're going to multiply the bottom one negative three so it becomes positive 9. i don't know why you're right positive 9 now here is going to be negative 9 over 7 cosine 3 t and then we divide by a negative 7 so it's going to be positive 3 over 7. so that's what we have
9:15:15
uh and then integrate okay so integrate
9:15:20
so once you integrate uh you're going to get the answer so y equal the integral of y prime dt so just integrate so you get negative 1 over 21 t cube negative 2 e to the negative t and then sine means uh negative cosine so is being 1 over 7 cosine t and then negative you divided by 3 so you get 3 over 7 sine 3t plus 3 over 7 you add another constant c2 for that so those are the two answers so we have the x and the y so you can see just very direct way to
9:16:02
simplify that so that's a type 1 so let's talk about the type two so what is type two
9:16:10
so type two um type two like uh the situation there we can do this okay so let's let's look at a problem again so let's say uh solve the system solve the system let's say x prime equal 3y y prime equal to x minus y plus three so that's the system sometimes we write the curly brace like that it's not really necessary uh we're gonna call the equations one and equation two now you can see uh there's no way you can just solve for one here because there are
9:16:53
other derivatives so that means the method that we used for the previous one doesn't work here so what you're going to do is you're going to take derivatives you can take derivatives and then try to substitute from the other one to remove one variable now you have two options you can take the derivative the first one or the derivative of the second one but you can see the three here since we have a three here to avoid undetermined uh coefficient method
9:17:20
we can take the derivative of the bottom one then that's three gonna go away so that's the that's the trick we do you see because if you can derive the top one you end up with the equation which is undetermined you have to use the undetermined coefficient method so that can be little difficult so we can avoid that take the derivative of the bottom one okay so we have things sometimes so take the derivative of the second equation so the answer uh taking the derivative
9:17:47
taking the derivative of our equation number two uh so you're gonna get y double prime equal to x prime minus y prime and then plus zero that's gonna go away now what we're gonna substitute for uh for x prime okay so substitute uh from substitute four let's say subject four uh substitute for x prime so if you do that you get y double prime equal two times what is x prime x prime is 3y from the first equation minus y prime not now what we can do we can bring
9:18:32
everything to one side so it is y double prime plus y prime minus 6 y equals 0 this is a simple constant coefficient differential equation and we write the characteristic equation for that so the characteristic equation the characteristic equation is lambda squared plus lambda minus six equals zero and which has factors uh lambda minus lambda plus three and lambda minus two equals zero this is gonna say that lambda equal negative 3 and 2 so that means you can
9:19:10
write the solution okay so the y is simply using the formula so y is simply c 1 e to the negative 3 t plus c 2 e to the 2 t so that is the y now what we can do um in this problem since we know y uh you can because the goal is to find x so what we can do is you can plug in that to one of those equations wherever you see y and then solve for x so you can see that you can do that in two different ways uh the one is you can just substitute for take the derivative
9:19:51
uh because we already know what is uh the derivative y prime and then we can substitute that uh take the derivative and plug into the equation number two then you get x or substitute in the one in the one um you know we know that these y values substitute there and then integrate again so we can use that one because that will look a little simple so by 1 are we going to say x prime equal 3y so now just plug in for y so it is 3 times c 1 e to the negative 3 t plus 3 times c to e to the 2 t now
9:20:28
what we do integrate okay so integrate
9:20:35
so you're gonna get x which is the integral of x prime d t just integrate the terms so that means you can divide by negative three um so you're going to get let's write 3c1 times e to the negative 3t so this step is not really so you can just write the answer directly and then here are the same thing so 3 c 2 e to the 2 t divided by 2 now there is a question whether we are going to add another constant and you can see that this is a we have two first order equations
9:21:18
uh that means you only need two constants we already have the constants so that means we are not gonna add anything new okay we are not gonna add c3 here so i'm gonna write but i'm going to say we are not gonna do that okay because we already have two constants so that's the reason we have c1 and c2 we only need two constants good and then we can simplify if you want so that means the final answer gonna be negative c1 e to the negative three t plus three halves ah c two e to the two t
9:21:55
yeah that's how we uh do it uh so you can see it's not that difficult but uh you have to kind of think before jump okay because which one take uh which one you take the derivative uh depend uh so you know can give you like a different answer i mean it can be a little lengthy if you take the derivative of the first one because you have so undetermined coefficient problem okay but if you take the data to the bottom one is much easier so it's always depend on the
9:22:19
problem you can't see beforehand what's gonna happen okay let's go to the next type next topic the laplace transform method so
9:22:30
laplace transform method so what we're going to do uh let's again look at a problem so what we normally do is so let's discuss the argument so we can take the laplace transform of the equation so because we can see it's a system now so you can see many equations so we take the laplace transform all the equations so take a laplace transform of uh the equations the equations instead of one equation and solve and so after you get the equations you're going to see a system with those uh xs ys
9:23:15
things like that so we can solve for them as we did before okay and solve for uh x s y s uh like that because we don't know how many we have so solve for them using the elimination method uh so you can see uh yeah let's write i'm gonna say field using elimination
9:23:46
then uh take the inverse laplace to find uh x t y t like this so it's a very simple idea take the laplace transform so you get a system in that system you can see it is just uh there are no derivatives you just use the elimination to find them and then take the inverse laplace so uh yeah so take the laplace transform eliminate and then take the inverse laplace transform those are the steps so let's just do one problem but i'm not going to completely solve the problem because
9:24:21
those can be a little lengthy i'm going to show you the steps uh and then you can finish this so let's do this problem so example
9:24:33
solve the system
9:24:39
let's say d squared x over d t squared plus 3 x minus y equals something and then let's say we have d y over d t this is a interesting part you can have anything here uh not like before uh so it is negative 4x plus 2y some numbers and then 3 e negative t with uh the one of the problem here you need to give the initial conditions because you can solve so let's say x 0 y x 0 x prime 0 and y 0 all of them are 0 just to make little bit easier okay so that's the problem now what we're going to do
9:25:26
plus transform those two equations okay so answer uh taking laplace transform taking laplace transform uh we're gonna have in the first one we're gonna have s squared x s plus three x s minus y is equal what's laplace transform four of s so we'll get that equation and you can see that just excess there are no derivatives no other powers from the second one you're going to get is ys minus 4 xs plus 2 ys equal 3 this is the 8th function so it is s plus 1 those are 2 equations that we get
9:26:17
so we can isolate the s x term as a factor so and then you can also combine these two so let's do that so if you combine them okay so combining let's say combining ah we have s squared plus 3 s x minus y s equal 4 over uh in the other one you can see um negative for s6 is this it's a good idea to write s6 first and then yx next so that you see the pattern the other one you're gonna see s plus two yeah i'm gonna write this little closer to that so it is plus uh s plus two
9:27:11
y s uh equal three over s plus one i'm going to call this equation 1 and call the equation 2. now think of how to remove how to eliminate one of the variables so for example you can see let's start eliminating the ys if you want to eliminate ys think of this term because that's the coefficient so what we can do we're going to match the coefficient with the opposite sign so for example if you multiply the first equation by s plus 2 the whole equation by s plus 2 and added
9:27:41
to the second equation the yes term gets away it's going to go away okay if you multiply the first equation by s plus 2 and add it to the second equation the y is term going to go away so you can solve for sx so that's the whole point so let's do it uh so by uh we can do s plus 2 times equation 1 plus 2. if you do that you can see the y s term has the same coefficient so when you add them same and opposite coefficients so when you add them they're going to cancel out so you simply have s plus 2
9:28:23
times s squared plus 3 minus 4 from the bottom one times sx that's what you have on the other side you have 4 times s plus 2 over s plus three times s plus one that's a ticket um so if you want you can simplify this in uh to get one expression or leave as two so how you want to do it now you uh notice something you can see that if you simplify uh the what is in the bracket you get s 3 plus 2 s square plus 3 s plus 2 and then which has factors is plus 1 and
9:29:17
s squared plus s plus 2. and if you don't know how to do this i'll look at the other video about fracturing polynomials so that explain like how to find factors like this so now what we're going to do we can do both sides by that so you're going to get sx so let's divide both sides so we have [Music] i mean if you had s plus 2 here then you could cancel but it doesn't happen like that so we get 4 [Music] 4 s plus 8 divided by s plus 1 s squared plus s plus 2 and also we have extra s in front
9:29:59
and plus 3 uh what's going to happen here you have s plus 1 already so it is s plus 2 squared s squared plus s plus 2. now what you want to do is just invert so use the uh you know try to find factors uh the partial fractions and invert so you can do either that way or also you can use the convolution uh if you want to inverse or whatever the method that you use what you want to do now is the inverting but i will not do that i will say okay so uh invert
9:30:37
to find
9:30:41
x t and then what do you do uh you can plug in this x t in one of those equations and solve for y s or do the same thing as before try to eliminate uh the s x term by just switching the variables and multiplying the opposite one and adding so you can use one of those methods to solve for y x and x t okay and then you can see uh sim uh so you can do an invert okay and then uh the next one is uh find ys similarly
9:31:24
to do that to do that uh substitute substitute x s because we already know is s x in in two and we can say solve for ys
9:31:50
and invert this just one way to do it the other way is just do the uh elimination as before and solve for you so it's it's sometimes i mean depend on what we have oh oh i use elimination to find a ys so you have two ways to do either you substitute uh into the other one because we already know uh by s we already know s6 and solve for ys or just use the elimination to find ys so you know it depends okay so that's how you finish the problems i'm not
9:32:32
gonna do it because it can take too much time but this is the way to do it uh because this is very useful when you have like a large system uh so laplace transform method gonna help you to find the solution so so we know the the method so let's move on to the the last one uh that is the eigenvector method so let's talk about that so eigenvector method are you can vector or you can either value method okay eigenvector method so let's talk about the the steps uh
9:33:09
so we're gonna go with the first step uh what we're gonna do we're gonna write the system uh using a matrix form so right the system
9:33:21
write the system in write the system in matrix form
9:33:31
that means x prime equal ax form that's the first step and then what we're going to find eigenvalues find eigenvalues eigen values lambda i of a that's a second step the third step is use any method use any method
9:34:01
so you can either use the shortcut that we discussed earlier so there are several videos for that for the 2x2 case and 3x3 case so use the shortcut or
9:34:15
shortcut or or the regular method the gauss elimination method okay the shortcut or the regular method to find to find the corresponding eigenvectors
9:34:36
corresponding
9:34:41
eigenvectors so you cannot use the shortcut or the regular limit so what card is really short okay so ah and then ah let's say v i i can vectors v s the corresponding vector and then we write the solution so write the solution ah so we know that the solution is x t which is given by c 1 the v 1 e lambda 1 t so the v 1 is the corresponding eigenvector of lambda 1 and then like just like that so we're going to go v2 e lambda 2 t so like that so there are n many sorry c n v n e lambda
9:35:29
n t also there are as not
9:35:36
so lambda i's may be uh equal
9:35:44
may be equal but the v i's v1 v2 v n should be should be linearly
9:36:06
independent linearly so that's the only requirement we need those eigenvectors has to be linearly independent but there's a fact
9:36:21
this we discussed earlier uh the eigenvectors of different eigenvalues are linearly independent always so eigenvectors
9:36:33
eigenvectors of different
9:36:42
eigenvalues
9:36:47
are linearly independent that you can show that's the case they are linearly independent but sometimes what happened even if you have some repeated eigen values uh you can show that those vectors are linearly independent in that case uh you can still write the solution in this form okay so we're going to use this form for two cases the distinct eigenvalues or repeated eigenvalues but with linearly independent eigenvectors okay so those are the two cases a one with distinct eigenvalues
9:37:23
or one with uh some equal eigenvalues repeat eigenvalues but with linearly independent eigenvectors so if you want we can like kind of repeat that because that's a kind of important step
9:37:41
so we use this form we use this form of the solution
9:37:53
when for a distinct eigenvalues for distinct eigenvalues or some repeated eigenvalues some repeated
9:38:15
eigen values with linearly independent with linearly independent eigenvectors
9:38:27
v1 v2 through vn so for those two cases uh we can use this form of the uh solution so some repeated eigen values with linearly independent is important in independent eigenvectors okay so let's try to do uh two examples one from two by two and another form uh uh three by three so we start with the two by two one so uh so example one so let's say use eigenvector method
9:39:02
eigenvector method this eigenvector method to solve
9:39:11
the system so the system uh let's say we have x prime equal 3x plus 4 3 x plus let's say 3x plus y and then y prime equal to x plus 4y so this is a system um so we can do this using the element as a type 2 elimination method you can take the derivative and plug in from the other one you can do that and then eliminate one of the variable uh
9:39:50
let's see how to use the eigenvector eigenvector also the answer so we're going to let x which is the vector of x and y those are the solutions ah then we know that x prime is simply x prime y prime so we can write the system in this form so uh so write the system in matrix form so the next one is right so i will write the steps but you don't want to do that i mean it's just guiding you but in the exam you don't write those things so write the system
9:40:31
you can to make it like more logical so write the system in matrix form
9:40:40
so we have x prime uh the coefficients three one two four so those are the coefficients times x so we just put the coefficient and put x there now what we do we're going to find the eigenvalues okay so we're going to so this is a so i'm going to call this is a so we're going to find eigen values of next step find
9:41:10
eigenvalues of a ah for that we can look at the determinant of a minus lambda i so that is the i subtract the lambda from the diagonal entry so it is 3 minus lambda 1 2 four minus lambda so this is simple so this is a two by two so we're gonna multiply the diagonal first so multiply the diagonal and subtract the other diagram so minus other diagonal means 2 you set this equal to 0 and you can see that this simplifies to lambda squared minus 7 lambda
9:41:56
plus 10 equals 0. this has factors lambda minus 2 and lambda minus 5 equals 0. this implies that lambda equal 2 and 5. so those are the eigenvalues okay uh so the eigen values are
9:42:21
lambda equal to m phi now what we can do we're going to find the corresponding uh eigenvectors so find
9:42:33
eigenvectors
9:42:38
find eigenvectors for that we can use a shortcut method that means uh eigenvectors are simply the columns of the uh complementary eigen matrix the other eigen matrix so remember the eigenvectors are the columns of non-zero columns of the other eigen matrix so we're going to do that so for lambda equal 2 we're going to calculate the kappa matrix of phi because that's the other eigenvalue so that means you subtract 5 from the diagonal 3 minus 5 1 2 4 minus 5 so you're going to get the columns as
9:43:23
negative 2 2 1 negative 1 these are the eigen vectors and you see that they are just a multiple of each other so that's the eigenvector so that means this implies that eigenvector for lambda equal to uh the column just pick any column or any multiple so one negative one that's eigenvector okay so you simply pick one of them or even a scale version of that so i pick that one or you can pick the other one or you can even multiply by something uh
9:43:56
now what we're going to do we can do the same thing with the other eigenvalues so for
9:44:03
lambda equals 5 now we're going to calculate the other one so it is 2 here now so if you use 2 subtract 2 so it is 3 minus 2 and 2 1 4 minus 2 so you're gonna get one two one two so you can see they are the same and that's the eigenvector four lambda five so which is uh it's eigenvector for lambda equals 5 so which is 1 2. so we find eigen vectors now what we can do uh just just use the formula okay so very simple so the formula so uh now write the solution so we can write
9:44:52
the solution so which is x t equals c 1 v e lambda 1 t plus c 2 v 2 e lambda 2 t so that means c 1 eigen vector of 2 is 1 negative 1 so e 2 t plus c 2 eigen vector of 5 is 1 2 so it is e to the 5 t so that's the solution also you can simplify if you want but that's not necessary uh i'm just gonna write it in a you may simplify
9:45:38
but that's not necessary so let's say how to simplify this one then you can write x t so what we're going to distribute this so that means you can first say c 1 negative c 1 times e to the 2 t and if you multiply by c 2 it is c 2 2 c 2 a to the 5 t now we're going to multiply inside and add so that means if you multiply inside and add you're going to get c 1 e to the 2 t from the first one from the second one it is c 2 e to the 5 t and here also it is negative
9:46:15
c 1 e to the 2 t plus 2 c 2 e to the 5 t that means those are the two solutions so uh so therefore you can say the x solution that's the first component which is c 1 e to the 2 t plus c 2 e to the 5 t and the y t that's the second component with us how we define the x ah so actually this is not simple x this is uh this is a vector x so this is the big red ah it's a vector solution yeah even here so that's the uh vector solution good so it is uh and then we have c 1 e to the 2 t plus 2
9:47:09
c 2 e to the phi t so that's the solution so you can write it like that or you can just leave the first version that's fine good so that's the service holder a system of 2x2 now let's try to do one from three by three so let's solve this one for example
9:47:30
uh yeah actually uh to be better we're gonna put everything in one side okay and then one more thing ah if this is just a not ah if initial conditions if initial conditions are given
9:47:55
are given we can find
9:48:01
we can find c1 and c2 because if you know the initial conditions just plug in them here and solve for c1 and c2 so it's a simple uh simple task okay so let's do the three by three now uh so the three by three let's do uh just one example just to remember uh recall what we did but there are like three uh four examples solved in the lecture so you can look at if you need more examples so let's say solve there are three uh solve the system
9:48:39
system of ods r sorry is dx over dt equal three x minus y minus z d y over d t which is x plus y minus z d z over d t uh which is x minus y plus c so what we can do are we gonna write the system in the matrix form that's the first step so the answer if it is not in the matrix form write in a matrix form so uh so we're going to define the x vector which has three components x y and z ah now what we're going to writing the system
9:49:38
in matrix form matrix form okay writing system in matrix form so we have x prime equal the coefficients 3 negative 1 negative 1 1 1 negative 1 1 negative 1 1 times x that's the system matrix a now what we're going to do we can find the eigenvalues first okay so find eigenvalues
9:50:14
find eigenvalues so the zero equals the determinant of a minus lambda i uh so that means you subtract the lambda from the diagonal entry so 3 minus lambda 1 1 negative 1 1 minus lambda negative 1 and then a negative 1 negative 1 1 minus lambda so that's certificate so you can use the laplace expansion uh or the cofactor expansion to find the get the equation but sometimes it may be a good idea to use the determinant properties to simplify this first like you know try to get
9:50:53
like more zeros you can see they are one one they are equal so if you subtract them that means if you add the negative one of the other row you get a zero so it's always a good idea to get extra zeros then it's the calculation gonna be easier so we can do that so the first thing are we gonna do is uh we notice that there's a zero here there's a um they are equal so we're gonna get a zero uh so we're gonna subtract so what we're gonna do is i'm gonna uh subscribe i'm gonna add the negative uh
9:51:23
uh 1 of row 2 to row 1 so let's do that so if you do that so that means you subtract row 2 from row 1 so you get 2 minus lambda and you get lambda minus 2. so they are equal just opposite sign and this is 0. so that means you can take the lambda minus 2 as a factor so that's a good thing ah and then here we have 1 and then 1 minus lambda negative 1 1 negative 1 1 minus lambda so how we did what we did we when r1 minus r2 now uh what we can do is we can pull the uh lambda minus
9:52:02
2 out as a factor so if you do that it is lambda minus 2 as a factor from the first row so you get minus 1 and 1 1 1 1 minus lambda minus 1 so we have a 0 here minus 1 1 minus 1 now what we can do is you can add row 1 or column 1 to column 2 to get another 0. so let's do that so we're gonna do that so we're gonna get one more zero uh by doing so i mean you don't have to do that but this could have made the calculation much easier because it's easy to find the factor so
9:52:42
what we're gonna do are we going to add column 1 to column 2. so it is negative 1 1 1 you get another 0 and 1 minus lambda now it's 2 minus lambda because you add so this is going to be 2 minus lambda you add column 1 to column 2. so this becomes 2 minus lambda this is 0 and then 0 minus 1 1 minus lambda so what we did we add uh column one to column two okay and now we can write the expansion so once you have the expansion you're gonna expand uh at this point and you can see you
9:53:23
only have one uh term you get so that means if you write the expansion now you're going to get uh it's a negative because negative 1 times lambda minus 2 the 2 by 2 determinant which is 2 minus 0 minus 1 there's a 0 here 1 minus lambda so that means you already get the factored version negative lambda minus 2 2 minus lambda 1 minus lambda okay so this is simply if you switch a term this is lambda minus 2 squared 1 minus lambda if you set this equal to 0
9:53:59
this says that lambda equal 1 and 2 2 so that's a repetition so that means there are only two eigen values uh so this is the easy case because there are only two eigenvalues what we normally do we're gonna uh find the um so let's write that two eigenvalues because that's exaggeration two eigenvalues so what we normally do when you have two eigenvalues we're going to multiply the corresponding eigen matrices ah so the find the eigenvectors find eigenvectors
9:54:36
so what we do since there are only two eigenvalues we're gonna uh i use uh eiger matrices
9:54:48
so we can use eigen matrices for that uh so what you're gonna get is so we're gonna um do the eigen matrix product so it's a one eigenvalue is one of a times the other uh so we just get the two copper matrices and we're gonna multiply them so let's do that so what is the first eigenvalue is 1 so we can subtract 1 from the diagonal so you get 2 1 1 and then negative 1 you subtract 1 so you get uh negative 1 and then negative 1 negative 1 and 0
9:55:21
that's the first one the second one is are you going to subtract 2 from the diagonal so you get 1 1 1 negative 1 negative 1 negative 1 negative 1 negative 1 negative now what we can do we're going to multiply we're going to multiply and see what you get if you get 0 you're lucky the columns are eigen vectors so you can see that for example if you multiply this row with this column you're going to get 0. you got 2 minus 1 minus 1 so 0 and you can see that everything is 0
9:55:52
okay very good that means the columns are the eigenvectors of the other eigenvalue okay so that means uh so these are these columns there are three columns here these columns are the eigen vectors of lambda equal 2 and these are the eigenvectors of lambda equal 1. so it's always the opposite one good solution eigenvectors uh so the eigenvectors are
9:56:33
eigenvectors ah for v lambda one you can see they are all the same so it is 1 1 or just multiple of that so you can see that's exactly you see that there and for lambda equal to uh we need two eigenvectors uh just pick any column you like so any column or even any linear combination or rescaling for example i can divide the second column by negative 1 so you get 1 0 1 and then if you divide the second third column by negative 1 you get 1 1 zero also there are more things for
9:57:11
example if you add the column one to column two uh you're gonna get one one so that's another one uh so you can do a lot of other things so you can just add them you can multiply by one something and add them okay just any column operation so but we need only three so and you can also see that those three are linearly independent okay so they are linearly independent just by observation because where the zero is when you have a 0 like that it has to be linearly independent okay so they are
9:57:43
linearly independent
9:57:51
this is the early independent equal of the solution ah so the solution uh vector x t which is simply c 1 v 1 e lambda 1 t c 2 v 2 e lambda 2 t c 3 v 3 e lambda 3 t so if you substitute you'll get c 1 times the first one 1 1 1 e to the lambda what's the lambda value is 1 so it's t and the second one is c 2 times 1 0 1 the lambda value is 2 t the third one is c 3 uh the eigenvector is 1 1 0 e to the 2 t so that is the solution of the system as before we can if you want you can just multiply
9:58:43
through and add them and then get 3 equations but that's not necessary but this is the solution ah there are c1 c2 and c3 are constants so you can see it's not that difficult it's very simple but this is for uh two eigenvalues if there are three eigenvalues what we normally do is we're gonna pick um like if you need the eigenvector for lambda one what we can do we're gonna pick the other two eigen matrices and just multiply them but we only need one column multiplied by the
9:59:21
other matrix so it's very easy you can get the answer really fast you only pick one column normally with minus zeros so we can multiply that with the other eigen matrix so you get eigen vector just keep doing it for other ones so you pick the other two matrices pick a column multiplied by the other matrix you get eigenvector so just simple multiplication so that's the uh end of this review
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