Skip to main content
Loading Explore guides…
First-Order Linear Differential Equations · Acedra
Home
Explore
Mathematics
First-Order Linear Differential Equations
Sign in to save
Chat
If YouTube asks you to sign in, open the video on YouTube.
Watch on YouTube
Reload player
Transcript
53 segments
Copy
Hide video
0:00
everything will work out nicely [Music]
0:07
in this video i'm going to talk about the idea and the strategy for how to solve a first-order linear differential equation but before i do this with you guys let's talk about an integral first let me ask you guys this what's the integral of sine x over x and i know i did not put down the dx but it was for good purpose anyways what is the answer to this
0:27
and you should know that this right here has no answer right this is actually a non-elementary integral so no matter what you do use substitution integration by parts uh whatever however the answer at the end is just not going to be nice right but what if i tell you that maybe i can add something right here let me ask you guys this following integra instead what if i'm going to ask you guys what's the integral of x over x plus l and x times cosine x
0:58
and now suppose this is my actual question i want to ask you guys i'll put a parenthesis around this and a dx so i can be legitimate what's the answer to this integral now
1:11
okay maybe you see it maybe not but let me tell you that this right here it actually has a much better answer and the answer to this integral is going to be sine x times l1 x and of course we are done and i'll put down a plus c at the end
1:29
well how come i can just get this for the answer well let's look at this backwards okay if you just kind of pay attention to the form first of all you see that you have a addition in between of two things right and then here we have sine x over x and then here we have l1 x times cosine x and guess what if you put this down you see sine x times ln x that's the function part and if you just go ahead and differentiate this instead this is how we do internet derivative
2:02
right you put down the answer and if you want to know if this is the right answer or not you just have to take the derivative of this if you can get back that would be the anti-derivative that would be the integral huh you have to use the product rule because this is a product of two functions and the product rule to show you this well i'll keep the first function which is sine x and then we multiply by the derivative of second which is going to be 1 over x and then we add and
2:27
you see this is why i know that i will use the product 2 function because earlier this is the adding situation right we add the second function which is l and x yeah that's a big hint you see
2:41
and we multiply by the derivative of the first function derivative of x is positive cosine x and you see we get back to the original integrand and of course this will be the anti-derivative for that and that would be the answer earlier huh so that will be a way to do integral right you just kind of look at the form and then going to just notice that this right here it actually came from
3:07
a product rule situation for the derivative okay now why did i bring this up let's talk about the linear differential equation if you look at this right here we have this is the formula you want to begin with d y d x which is the derivative plus another function times y and y is a function of x and then you put everything else on the right hand side and you see this is linear because the
3:34
derivative okay it's just to the first power it's not the cosine it's not being squared whatsoever likewise for the y it's only to the first power if you have y squared if you have square root of y this wouldn't work okay so i'm going to show you guys a strategy of how to deal with this kind of situation look at this again we have the derivative and the original [Music] and i'm going to use this idea again let me write it down for you guys here is the product rule i just
4:06
want to put this down and i would like to write it down as if you want to take the derivative of two functions and i'll just use f times g f is the first and g is the second of course and the form i want to write it down for you guys is this we will keep the first function and we multiply by the derivative second and then we add the second function times the derivative of the first this is the product rule right and now i'm going to differentiate
4:35
this for you guys you see we have the derivative and it's original we have the derivative well in this case it would be the g prime if you want to match like right here and the original which is right here very similar right however over there we have an issue because there was nothing in front and of course it's not the case that you have a one in front that gets a function wise
5:01
anyways i want to just kind of match the phone for you guys today i want to differentiate something times y and i'm not sure what this is yet because i didn't have anything right here so that's why i just leave it blank okay all right i want to differentiate something times y you can imagine this actually a function okay all right using the product rule we will have something times the derivative of the second function which i'll just put down the y dx so
5:33
i can match with that form over there and then we add well this right here i don't know what it is yet and i'll just kind of like i don't know b for blank but then the second part is we will keep the second function which is y times whatever this was and now you see this is why i put down
5:55
for the linear differential equation i'm looking at the product rule once again i didn't have anything in the front so that's why i leave a space right here and then for this right here it's the one i have to differentiate all right now the issue is seriously we didn't have anything in the front and this may not be a good phone that we want to use so this is the idea and the strategy
6:23
we're going to use something now we can force the left hand side here to be the derivative that came from the product of two functions and what we're going to do right here is that we'll just say we're going to use and the thing is that we will multiply what we mean by the inter grading factor
6:49
okay so we can force it the left-hand side to be the product of two functions and we have to use the product rule to match that well we are going to use another function and that's called the integrating factor and i will label this as mu of x okay so i'm going to kind of just draw a line between this and this is what we're going to do we will multiply everything by mu of x so from here
7:18
we will end up mu of x times d y d x plus mu of x times p of x times y and then on the right hand side we have to just multiply this and that and i'll just have mu of x times q of x just like that
7:39
all right now you see i'm just going to put on mu of x right that's a new thing i want to have otherwise seriously i just have blanks that's no fun right so the blank will be the mu of x and right here let me just write down mu instead of mu of x for simplicity purpose hopefully you guys don't mind today if we differentiate mu which is a function of x times y which is also the function of x we use the product rule we keep the first function which is mu and then for the second part
8:12
right here you see we will have to differentiate the mu and then put it here and now you see
8:19
we can mix and match the form mu of x times dydx which is this right here and then we add the words we have the y right here that's matched right
8:31
well in order for us to come up with a formula for mu of x what must happen is you see mu of x times
8:39
p of x which p of x was from the original it must match mu prime so the key is we are going to set this to be mu prime and let me write this down right here for you guys mu of x times p of x we want this to be so let's set this to be mu prime of x because this way you will see the left hand side will be the derivative of a product of two functions right and from this right here we'll be able to come up with a formula to get mu of x and we will multiply everything by
9:15
the mu of x the actual function and you'll see everything will work out nicely of course i'll show you guys an example with actual functions so be sure to continue to watch this video and also i would suggest you guys to watch another video it's called the fake product rule you can click the link in the description it's a lot of fun and the strategy right here is very similar okay
9:36
now let's continue what i want to do next is i would like to divide both sides by mu of x okay so that they can so and let me write this down first right here for you guys we will have mu prime of x over mu of x and this is equal to p of x and now let's look at the left hand side we have the original function in the denominator and the derivative of the numerator
10:06
this is a form that came from the derivative of a logarithm right in another word if i integrate this and the reason i'm integrating is so that i can somehow get rid of the derivative right anyways let me integrate this of course we have to integrate both sides
10:25
this right here is going to be ln absolute value of mu of x we all double check today if we differentiate this the derivative of ln of mu of x is going to be 1 over the inside 1 over the mu of x and then because of the chain rule we multiply by the derivative of the inside that's how we get the mean prime of x okay so the integral of this is equal to ln of this if you differentiate that you get that don't worry about the plus z and on the right hand side i will
11:01
just write it down as how it is integral of p of x dx and once again don't worry about the plus z because we don't care about the constant we just care about the function part we just care about that function part so we can multiply the original equation by the function part everything will work out nicely at the end we see that we have mu of x instead of the ln
11:26
what can we do to get rid of the ln we do e to that e to that right so that e and now cancel finally this is the formula and let me write this down right here for you guys to get mu of x
11:43
and you don't even need to worry about the absolute value because once again you can have positive and one negative function both of them will work you just need to have one function that you need to make things work so let me just write down mu of x this is equal to e okay and the power is an integral you have to integrate p of x dx this is how you are going to find out
12:11
what that integrating factor is and now let me show you guys an example so everything will make much more sense so let me show you guys how we can solve this linear differential equation with the strategy that we just talked about it and if you're doing this on your own you should pay attention to the following we see that we have the y dx the first derivative right and here we have the y to the first power there's no other y's there's no other derivative
12:36
this is indeed the first order linear differential equation so you can use the integrating factor the one i just showed you guys earlier right now this right here it's not in the correct form though because we have this cosine of x in front but it's okay because we can just go ahead and divide everything by cosine of x
13:00
right this way we'll be able to get into the correct form that we want we will have d y d x plus sine of x over cosine of x we can just write that down as tangent of x and you see right here i put a parenthesis around the x because we have to emphasize x is the only thing that's inside of the tangent the y is being multiplied it with tangent of x and this is equal to 1 over cosine of x which is just secant of x right so let's just put that down from here
13:36
we are in business because we can see that the p of x this right here is positive tangent of x
13:47
from here we will be able to work out the formula to find us ourselves an integrating factor and yes you have to know the formula so right here let me write it down mu of x for that integrating factor it's going to be e as the base and the power is an integral right we have to integrate p of x and let me just write this down for you guys we have to integrate tangent of x right here in this situation and let's just go ahead and work things out first work out the integral
14:21
the integral of tangent of x is ln absolute value of secant x right so we still have the e for the base but this is going to give us ln absolute value of secant of x
14:35
don't worry about putting down a plus c right here or not because we just need one integrating factor with just that function part the plus c later on yes it would be from the process of solving the differential equation but for the integrating factor don't worry about it and in fact what i'm going to do next is i would just like to cancel the e and the ln and i'm just going to say mu of
15:02
x instead of saying absolute value of secant x i'm just going to use the positive version secant of x this is it okay and the reason for that is we just once again we just need to have one function
15:17
you don't want to multiply with absolute value of secant x you don't want to multiply by negative nobody that's negative use the positive version positive secant of x this right here is a function that we're going to multiply throughout with our equation everything will work out nicely and i know i didn't put on the parentheses right here around here but they shouldn't body too much right anyways let's just multiply by secant of x throughout this equation and
15:44
let me write this down right here for you guys you multiply everything by secant of x
15:50
and of course we have to be sure to distribute distribute this trivial and let me put down the result right here first we'll have secant of x times dydx and let me write it down right here
16:06
next we add the words secant of x times tangent x right so let me put this down secant of x times tangent of x and then we still have this y right here and on the right hand side secant of x times secant of x we have secant square of x all right here is the punch line
16:31
on the left hand side this you see we have secant of x times the derivative of y plus y times the derivative of the secant x right this right here is nothing but just the derivative of secant x
16:52
times y you see if you go back to the phone earlier this was the mu of x and the y
17:00
and once again you can always write down the derivative and just differentiate it to double check derivative of secant x times y is you keep the secant x and times the derivative of y and then you add you keep the y right here and you multiply by the derivative secant x which is secant x tangent x so nice and you see this right here is just that we force it to be the derivative of the product of this and that the right hand side is still the secant square of x okay and the
17:31
left-hand side we have a derivative how can we get rid of this derivative well we can just integrate right and of course we will have to integrate both sides on the left-hand side this and that will cancel and we just have this right so let's write down secant of x times y i know this looks sexy but this is secant of x times y and then this is equal to the integral of secant square of x this
17:56
is just tangent of x and then right here i will put down the plus c and this is because you know we are solving that differential equation this is the constant that we need all right now at the end of course we want to isolate the y it's not bad to do right but let me just write this as 1 over cosine of x first and this is multiplied by y and this is equal to tangent of x let me just write
18:23
it as sum of x over cosine of x and we still have this plus c because if i look at the equation this 4 now we can just multiply everything by cosine of x this and that cancel out and we will just have the y left on the left hand side and then be sure you have to distribute right multiply everything by cosine of x cosine of x times this will just have sine of x and then cosine of x times plus
18:55
c we will just put down plus c times cosine of x and this right here is the general solution to that differential equation and of course suppose you have initial value let's say if you have this equation here along with an initial value that say y of pi is equal to two well now we can just plug in these values into this four and so for c this means x is equal to pi and y will be 2. now let's just plug in 2 into this y we'll have 2 this is equal to
19:33
sine of pi right because x is pi so let's write that down pi plus c we don't know yet and we multiply by cosine of x which is the same as cosine pi and just work this out real quick sine of pi is 0 cosine of pi is negative 1. so this means we have the 2 right here and this is just 0 and we have just c times negative 1 that's negative c and of course we can just divide both sides by negative 1. c will give us negative 2. at the end
20:09
if you have this initial value the solution will be y is equal to sine of x
20:16
and here we have the plus c but you know c is negative 2 so we'll have minus 2 times cosine of x and this right here will be the answer with this initial condition so hopefully you guys like this
Here to help you learn
Sign in to keep a private conversation with your course tutor.
Ask me anything about the material…
Sign in to chat