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0:41
>> Math for Electronics I Ma1101: All right. So let's begin. Good morning once again. So, today we will do a week 12 Okay, so week to always all about ordinary, differential equations.
0:52
Before I proceed. So this is the last week. oh,
1:00
in this particular week, we primarily spend to give you some Basic. Understanding on ordinary differential equation. So this water what is differential equation and what is First order linear differential equation. And second order. Linear. Differential equation ordinary differential equation. Okay, so majorly these two are the important things. That you have to definitely learn in Math, 1 at least and later on higher order differential equations are also really important and
1:31
that you will see in the future courses, but as far as this course concerned, We will majorly. Focus on. First order ordinary differential equation.
1:49
And the second order homogeneous constant coefficients ordinary differential equations. Okay, so That's the thing. Let's begin with the differential equation. What is differential? Equation mean. Okay, so you have a function assume that you have a function f, And a differential equation. Is an equation. It's basically an equation that involves. The function, f. Okay, along with its derivatives. okay, so along So it will have f within it.
2:31
and also f prime, which is the first derivative f double prime, which is the second derivative. And so on so forth, it can have Third, fourth, fifth and so on. okay, and other functions of X also need not be
2:48
A function of F, You can have some other functions like g of x p of x, some other functions also. Okay. so combination of all these things together, forming an equation is called the differential equation. And that equation. Need to be satisfied by. All x. Okay. So which it should be satisfied for all applicable values of X but in general, we
3:18
For will say for all x except few values. Okay, so that we will see For example. You see this F prime? F Prime of x is equals to 4. so this contains the faster derivative With respect to x. On the left hand side, and on the right hand side, you have a constant. Right, so this is a differential equation. And this equation. Is the first order.
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First order, differential equation. Why? Because it has only The first derivative. okay, so order actually represents The. okay, so here you have only Prime.
4:14
Okay. F prime. The first first order first derivative you have. So it is called first order delivery. Differential equation. Even the second example. Like f prime of x plus, f of x is equal to 2. Here. Also, you have the highest Derivative that you have is F prime of X. Right? Which is d f by dx, which is the first derivative Therefore this equation is also. Called. The first order already.
4:46
Okay so
4:52
Order is really important here. Okay, order is not what you think. Like it's not a sequence or something kept in order or something order is basically Represent the highest derivative function existing in the equation. okay, so now with that we can actually Say which kind of ordinary differential equation. Is this particular function is so it has f double prime of x. It has F prime of X. It has half of X. It also has sine x, which is another function of x, right? So, sine x
5:25
is some other function of x. Right.
5:34
So it contains all F of X and it also contains F prime and F double prime. Right. So here if you see Derivative. that is present in this particular equation is second derivative Right. So therefore this equation. Is. Classified as or called as second order. Ordinary differential equation. Okay, so This term order is really important to understand.
6:08
okay, so let me also Point that. it's first order because it has
6:20
It is second order because the highest power is to derivative is after it. Okay, I hope. The definition of differential. Equation is clear here. And this differential equations, actually.
6:33
You will see them in most of the physical world examples. For example.
6:41
You have a falling object with some air resistance. Okay. So you are you are in the second floor and you you're throwing some apple or you're dropping an apple from second floor to the ground floor. Okay, so what do you do? You just drop it? So, initially your apple might be here. But it will.
7:01
Slowly start dropping while it drops, right? So even in real world, you see right? Initially it will not have any speed. But as it falls, the speed will increase, right? So it has some acceleration. Attached to it, right? And acceleration is Derivative of the speed. Right. So We have. Some acceleration attached to it. And we know the initial speed will be 0. It is assumed to be 0. Right. So initially, you catch the apple and then just drop it. So initial speed is 0 and as time
7:37
progresses, the speed will increase And at a particular time, T. The position of the apple. Can be referred to as x of t. It depends on the time D.
7:53
And the velocity or the speed at that particular time, t's V of T. And we know the connection between the speed and the distance it travels, right? So the connection is v of t is
8:06
Dx by dt. So the distance traveled by the time, gives me the speed Right? So v of T is x prime of t. And we also know the prime of T is the acceleration. Right. So v prime of T, is some Which is the acceleration. We call it as some. A. Okay. So A or A of T. whichever it is, but we know we prime of T's The acceleration. now, what happens if you see at a particular time D, What all forces that act on this apple? so, there will be some air resistance because the object is falling downwards
8:56
now. The air will resist or it will oppose from the opposite direction. And it will be proportional to the speed. So, the air resistance will be k times u of T. and as the object is falling, it has some mass as it is falling downward, it has some Force, which is mg. Which is weight acting downwards, and also it has a speed V of T downwards. Right now using this neutral loss and everything. So we have G is basically the acceleration, it has right? So, Which is we prime of T.
9:38
Right? So the force is Ma and A is V. Prime of T.
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Right? So instead of A we write it as we prime of T, so the force The total force is basically. M into V prime of K and what all forces that are there.
9:56
One is kvt on the opposite direction. One is mg on the downward direction itself. since the object is moving in this direction, the forces that act
10:07
In the same direction is taken as positive and the force of the tact opposite to direction. The object moves is taken to be negative. Okay, so the sum of these two forces will be mg minus. kvt will be the total force acting on the object. right, the total force is Ma And where A is V. Prime of t.
10:33
So the M into V prime of T is equal to is actually equal to mg minus K v. okay, so if the if the object is thrown upwards Now there will be slightly differences, right? So if it is thrown upwards the object is moving. In the upward direction. Right mg will act. Downwards. Air resistance will also act downwards. Right, because object is moving upward. And air resistant will act. Opposite to the direction that the Object moves. so if it is, if the object is going upward
11:10
Kvt acts downwards mg also act downwards.
11:16
Okay, so then it will just be the sum mg. Plus kvt will be the total force acting on the object and it is acted. Again is to the direction of the Object, so it will be taken minus. okay, so you can you can think of You can start writing equations for when the object moves upwards. So here the equations are downward okay, so once you have this equation, We will just rewrite, this. And bring it to our form. Right? So M, we prime D, so you do some manipulations, you divide both
11:50
sides by M. And bring. Minus kvt to the left side.
12:00
Then we will have v prime of T plus k by m. V of t is equals to g. So, this is the differential equation with respect to V of T. Right. Okay. And the initial condition. This is called initial condition. Initial conditions are like before. The process begin. Okay, before the experiment begins.
12:32
We know something about the velocity. So here velocity is our variable. Okay, so velocity is the function for us. So we know something about the function, initially the velocity is 0. The speed is 0, So therefore at time, t is equal to 0, v of 0 is 0. So this is called initial condition.
12:54
And I will let you know why we will use this initial conditions. So in order to find the constants, we'll use this initial condition, but the most important thing is forming this Ordinary differential equation. Okay, so this is a
13:11
Real world example and we have just used all the forces that are acting on it.
13:18
We just use neutral loss basically. And we have came up with this first order. Linear ordinary differential equations. We need out here.
13:32
This is okay. Even in the lecture it is clearly explained. Okay. So Let me go ahead. So that is one example and we will solve that we will solve that
13:45
order. Differential equations in some time. Okay? But before that letter, let me also talk about this RC circuit with Constant voltage or DC source. Okay, so you have 5 volts. Attached to the circuit. Now, you have only resistor and the capacitor. Okay. So now, we will use the kit jobs. Voltage loss. Loss. Okay. So We know the total voltage across.
14:20
Resistor and the capacitor is equal to 5. And the voltage drop.
14:30
With respect to resistor is equals to r into i of t. Right? Where I oft is the current flowing through the circuit at time d is equals at at time t. Okay. And With respect to capacitor, I oft is equal to Yeah, it's a math class.
14:51
Am I going towards electrical or what?
14:59
Yeah, this is one application of math. Okay, so
15:06
Odie actually is definitely used in most of your Future coming courses. Okay? So you will see like you have to sit and solo ordinary differential equations. Okay. alright, so Let's keep fun apart and let's come back.
15:29
Okay, so using the circuit. Last, we will have these equations and we'll use this equations to form. This equation. Right. So, how will we find this? Equation.
15:45
We will derivate this first equation. Okay, so if you derivate this first equation with respect to time d, you'll have vr prime T. Plus we see Prime D. Is equal to 0. Okay, so we see.
16:06
Theory of T's first derivative is 1 by C, I of T. We can write it as 1 by C I of T. V hours of VR of T voltage across resistance.
16:20
First derivative, you have to derivate this expression. So if you derivate this, you'll have r. i prime of t on the Right hand side.
16:33
Right. So we are prime. T is R. I Prime T. So this will be r. I prime T. Plus 1 by C I of T. Is equal to 0.
16:48
okay, so now divide the whole equation by R, you will have
16:55
I prime of t is equal. I prime or t plus 1 by rc into, i of t, is equal to 0. we all know what is the initial current
17:06
Anybody. Using your electrical. Vstc knowledge. You have this sketch of last red. So, initially current will only flow or will only be in the resistor. And I of zero will simply be.
17:35
By R. Right. So
17:44
So that we will see. Okay but this is pretty trivial. It should be trivial for you guys, at least. Okay. So, we have the initial condition with us. We have the ordinary differential equation, formed with respect to y of t. Now current, if I solve this ordinary differential equation, the solution that I get is nothing but the current with respect to timed Okay, so current will be the solution to the different league. okay, so these are Few Real World examples. now, let's see how to
18:18
solve this first order linear OD is Okay, so it's general form will be like this. so, you will have F prime of X. Because it's first order will only have F prime of x f of x and some other functions of x. That's it. You'll not have any higher order derivatives. Okay, so we'll have f prime of X. Plus. Some function p of x multiplied by F of x.
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Is equal to some other function q of x, where p of x and q of x are known to us. So these are known to us.
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Okay? So the only unknown is f of x. F of x. And we are going to solve this OD and we'll get
19:06
And equation in terms of f of x and x. That's it, okay? So that will be the solution. So that is what f of x to be. This is all But p of x and q of X are given to us. Just now I have shown you two examples, right? So in falling object with air resistance, We have got this kind of expression.
19:27
V prime of T plus k by m into v of T is equal to G. Right. If you compare this with the general form, you will see that your p of T. Right. So here variable is T. Here it is, written x. Okay, don't get confused. So here P of T will be k by m. Q of T, will be whatever there in the right hand side, which is Right? So we have p of t and q of T.
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Similarly, RCS are given RC argued. We have, i prime of T plus 1 by r c, i of T. is equals to, It should be 0. Okay. It is zero actually. So here.
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I of t, i prime of T plus 1 by r c. I oft is equal to 0 here. If you compare again with the general form, you will see that p of T will be one by r c and q of t will be 0.
20:34
okay, so I kind of skip this not linear case but you can watch the lectures and understand The only linear case also. So these two are linear. And what is the difference between linear and nonlinear? the only difference is you will have you don't have any powers attached to any of your
20:55
Functions. Okay. So, f of x or f prime of x, or f, double prime of x. It will not have any powers. I mean, you will not have expressions like F of X whole square. Or f prime of x whole square. Or whole cube. okay, so all the power
21:14
For your function as well as your derivatives will be 1.
21:20
Okay, so those if you have your equation in that form, they are called linear. if you don't have that form there non-linear, okay? So here, if you see an off t gets multiplied with an off t, and you will have n square of t. Which makes it non-linear.
21:35
Okay, so most of our problems or questions in great assignments or practices, assignments, even in the final exam. Also, we will focus on Linear odious. Okay, so you can kind of skip this part. But it good to understand what is non-linear means. Okay, so now let's see.
21:56
How to find the solution to the first order linear Audi. So once I have the Differential equation in front of me. How do I get to the solution?
22:08
Okay, so let's see that. Let's start with the easy case. Let's say I have a differential equation where my p of x is becoming 0 or it is 0. okay, if my p of x is 0, How will my expression look for p of x is 0? I will not have this. Right. I'll only have F prime of X. Is equal to q of x, this thing. And you know, if you have d f by Dx is equal to something. what will be the solution you just integrate on both sides or the solution will just simply be the anti derivative of q of X?
22:45
Right? So if a frame of x is q of x. F of x will be the antiderivative of q of X. Plus some constancy integral constancy. Okay, so it's it will be as easy as that if your p of x is 0. Okay, so, our solution will simply be F of X is equal to
23:04
And the derivative of q of x. So, let me write it clearly. It will be F of x.
23:11
A q of x, okay? Which is a q of x is the angry derivative of q of X. Bless some constances. Okay, for example, on the right hand side, you can see if You have. your differential equation to be like this F. Prime of X is equal to sine omega naughty. What will be F of X? I just have to integrate on both sides, right? So if I have if I integrate on both sides, I know so sine Omega, not t, will be. One way, Omega not into minus cos Omega naught T. Plus some constant. So that
23:41
will be the solution. And this is unknown to us, right? So this will, this can take any value. Right? So it will not be unique. So for uniqueness, what we will do is we'll use the initial conditions. So we'll use the initial conditions and solve this C. And replace that C with the value that you get. Okay, so this is about easy case. What if? The next case slightly difficult. What if it is just a constant.
24:13
Which is independent of x. Okay? So now the expression will look like this F prime of x. Plus P into F of x is equal to Q of X. Where p is just a constant, which is independent of X. now, what you do is You try to make it like an easy case, what is easy case? we'll have something derivative is equals to so, some f prime of x, or Sorry. Some F of X. Some functions derivative is equal to some other function. So that's the easiest. So we'll try try to convert into the K's in order to do that.
24:55
This is just over over years, we just found it. Okay, so now we can use it how we did this now. I can't explain, but what you have to do is you just have to multiply E Power px on both sides. Okay. And we know that e power p x is never 0 because it's an exponential function, it will never busy Right, so if you multiply what will happen.
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It you will have on the left hand side. E, power p, x into F prime of x plus P into E. Power p x into F of x right? On the right hand side.
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For a second. You just keep it aside. You just focus on the left hand side. How does it looking?
25:40
E Power px into F prime of X plus P into E, power p x into F of X. So you are taking one function, derivating the second function. Plus Derivating the first function and you are taking the second function as it is. Okay, so it is looking like if I treat E power p x, as Some G of X. So, if you let this E, power p of X to be G of X. Then the left hand side, if you carefully observe, it will just be g of x into F prime of x. Plus G prime of x, into F of X. Does this expression sound
26:17
familiar? To us. So how does it looking? Is it the products derivative? If this derivative, >> 24F2100084 SUDHANSHU KUMAR: Certificate like that. >> Math for Electronics I Ma1101: Right. So now Can I replace this with simply G of X, into F of X? Derivative Where g of x is e. Power p x, right? So that is, what is done in this
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particular step. So we multiplied e power px so that the left hand side will become a new function, which is the product function which is e, power p, x into f of x. It's whole derivative. So something inside some function inside Derivative on the right hand side. You have E power P x into Q of X. Right? Which is like this our earlier case. Left hand side some functions derivative. Is equal to some other function. So if you have this kind of expression, we know it sounds right? So its answer
27:15
will be The left hand side, function will just be the anti derivative of the right hand side function, plus some constants. Right, we are just using that. So, if I have this, Then f of x will be and derivative of q of X plus C.
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Here your F of x is E. Power p x into F of x, okay? Trade, it like that. So the whole thing will be F of X. No,
27:45
So that will be equals to the antiderivative of. So everything is in the subscript, okay. so E, power P of Sorry, not everything Axis. So this is in the subscript. Okay, e power p x of q is one thing.
28:06
it's Andy Derivative,
28:12
okay. So that is what the right hand side, whatever there in the right hand side, you are you just have to take it integral.
28:18
Okay. And add a constancy that will be your answer. Now, what will have is E Power p. X into F of x is antiderivative.
28:30
Of this E power P x of Q p x into Q plus C. Now, what you do is you divide by E power p x on both sets. so if you divide by E, power p, x on both sides, you will have This solution. finally, we will have At 4x is equal to E. Power minus PX. And the derivative of E power p, x into Q. Plus C into E power minus px. Where p is known to us. P is known. Q is known. Therefore, I can find this antiderivative. And I know this expression, I know this expression. The only are known left is C.
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Okay, the C is unknown and we will find the C using the initial conditions. Okay. So now let's do an example. The example is whatever the example that I have discussed earlier the following object with air resistance. So we have this part, this particular expression, right? So we prime of t, plus k by m. V of t is equal to g. So, here p is.
29:38
A by M, which is independent of t, right? P is just a constant k by m. K is known. M is known. It is just a constant. Q of T is also constant. That is secondary, but The primary p p is constant for us. P of x is constant for us. Okay. So now what is the solution?
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What I have to do, I have right hand side function, I have to take, it's anti derivative right, so right side I have
30:12
Q of T to be G right side. I will have E power p, x into q of T, right? So e power Pt. Into q of T, is G. And p is k by M. If you substitute everything.
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What you will get is e, power p t. Q of T is E power k, t by M into G. Now, what I have to do, I have to find the integral of this.
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D is constant. Anyway, I just have to find the integral of E power K, T by M. Right? If I do the integral, I will have
30:53
Into e power k, t by m. Remember E, power a x antiderivative is 1 by a E power, A x. Okay. So, here you have K by M as A
31:12
So it will just become 1 by a so m by k. So m by k into g. E power k, t by m will be the antiderivative. once you have this antiderivative, We will multiply it with E, power minus Pt.
31:28
Right. So that that is what it is, right? So E, power minus p t into plus C into E, power minus B. So, if you substitute, you will finally have MG by K. Plus C into E, power minus k, t by m. How is that? You just have to substitute. Okay, so we have this handed derivative with us. So now the solution is v of T is E Power minus. K t by m into this antiderivative. Which is m g by k into e power k, t by m.
32:07
Plus C into E power minus p, t, which is minus k t by m. okay, now if you see This, and this. e power minus of something which is getting product with E power, the same something The powers will get added up and if you add them up it will just become zero. E Power 0 is 1. So this will just vanish You'll have only MG by K left.
32:35
So that is what it is mg by K plus C times E power minus k t by m.
32:42
Everything is known mgk. Everything is known. Except this constant C. For this constancy, what we will do is we will use this initial condition we of the v of 0 is equal to 0 so if you substitute in this particular equation, T is equal to 0 and v of 0 is equal to 0. You will finally get that C to be minus mg by K.
33:08
now, what you will do, you just substitute see with minus mg by K and you can finally write the expression. For v of t just in terms of t. Okay, so it will be v of T is equals to m g by k.
33:24
You just pull out mg by K, common, you'll have 1 minus E, power minus K, t by M. Okay. So just An exponential. Decay, but there is minus. So there will be a Increase. If you plot it, you will see that. At t is equal to 0. You will have t is equal to 0 1 minus 1, which is 0. So v of 0 is 0.
33:54
As time increases. Our velocity will also increase speed will also increase.
34:04
But will it go to infinite? Is the question or the question is as time goes to infinite. Will my v of T? Will also go to infinite. If you see from the expression, If you tend T to infinite. You will see that this this particular expression will go to 0 and you will be left with mg by k. so therefore what it means is as time, even though the time goes to infinite
34:32
My speed will not go beyond mg by K, so it will just converge to mg by K. Right. And in the time intervals of m by case, if you see You will see that.
34:45
It will attain 63% of its final value in the first M by k interval itself. Okay, so it will be 0.63 into mg by K. and in the second time interval, it will reach the 86% of its Final value. And in the third time interval, it will reach its 95% of its time while time value, right? The percent of its max value.
35:11
and from there, it will just convert slowly even though the time interval goes on to infinite, It will not go beyond. Mg break. So, that's it. So initially The speed difference will be too high. But as time progresses, the speed will slowly converge to MG by K.
35:30
okay, so from the plot also, you can understand and questions can also be slightly trickier if they ask you. To find.
35:43
How much percentage of its final value has been attained? Right. So for that, you need to know this plot. Okay, or even if you substitute T is equal to M by K. You can get it if you substitute, T is equal to M by K. It will be 1, minus E power minus 1. 1, minus E, power minus 1, will come out to be 0.63, okay? So from expression, also analytically, you can get to the answer, but if you have a rough idea on the plot, If it is mcq, then you can fight over choose from the options. okay,
36:17
so I hope this Case, which is P of x. Being a constant, is clear. No. Without.
36:33
Anyone. Easy case. And The case. Where p of x is constant. Okay. Now, let's see. this RC circuit example, also, To get much clarity. Okay. So the differential equation formed with this rc is argued, is i of t? I prime of t plus 1 by r c. I oft is equal to 0 so here we can see p of T is 1 by RC Which is this? And q of t is 0. Right.
37:16
And we first find EPA E. E power P t's into Q of T's. Anti derivative.
37:25
So e power p t into Q of T. Here Q of T is 0. So e power pt of q, e power p. T q of T will also be zero and the antiderivative of 0 is 0 so you will not have the first expression. Right. So first expression I mean, I of T. Is.
37:45
E power minus p t into antiderivative of E power B, t q. Of X. Right. So this is the solution right? First order. Linear ordinary. Differential equation solution. Once our Antiderivative is anti derivative is 0. This whole thing will become 0.
38:08
So our eye off table simply be c into E. Power minus p t and p is 1 by r c. So i
38:14
of T will simply be see into e power minus T by r c that's the solution. The only Unown here is again. See But we know the initial condition initial condition will be, i of 0. is just
38:29
This so 5 is equal to r into. I of i So is Phi by R. So from there you get, I of 0 is 5 by R. So, i of 0, if I substitute. In this t is equal to 0 if I substitute in this particular equation I'll have I of 0 is equals to c. I of 0 is 5 by R, so C will be 5 by R.
38:53
I substitute that phi by R in place of C. Finally. I will have my solution to the ordinary differential equation.
39:02
Which is i of t is equal to 5 by r into e power minus T by r c. Even this also, you can plot. and this is, Here. Also, you will observe that Initially, the current is 5 by R.
39:22
But as the time progress, the current drops exponentially. In the first. Or in the first RC seconds. We will drop by. Right. So and then,
39:47
IT WILL DROP BY 63% SO IT WILL DROP TO 0.37 37 PERCENT of the INITIAL value.
39:53
and in the first two time constants, so this is rc is called the time constant before. Also m m by k is the time constant Okay, so We will talk in faster time constant 2 time constants three times constants like that. okay, so, in three time constants, It will drop to 5% of the initial value. okay, and as time Goes to infinite. If you just leave,
40:23
It the current will slowly come towards zero, so finally there will not be any current at all. Okay. So that is the plot. Okay, any doubts in first order?
40:37
Anybody. Okay. So the last 15 minutes, let's talk about second order linear, Audi is also. Okay and mostly in second order linear audience.
40:55
We will primarily focus on the constant coefficients. So instead of p of x q of x and everything, we will have all p of x q of x.
41:06
The left hand side's coefficients will be constant, will read the mass constant. The right hand side may be a function, but the left hand side. Will everything be a constant? Okay, again. It's all of your electrical engineer's.
41:24
You would have also studied about this RLC circuit, right? So KCl Kvl Right. Are we all aware of this? Kirchoffs. Circuit class voltage loss.
41:44
Last. Yeah, okay, so if you know that then you are just using those laws here. that's it. So here we have resistor inductor capacitor. So, all three are there in a series. They are connected in a series. Batteries attached to this. Okay, so there are two cases, DC, source and AC source.
42:08
DC is just the constant value. AC. You can just be a sinusoidal, okay. It can be a sign. Omega T or a cosm of 30, whatever. Whichever it is. Okay, so once it is attached. And if you turn the switch on, Then the current starts flowing. Right? And you have the circuit last with us, so We are. Should be equals to vs. Right. So that is
42:41
The first law and each, and every component will have its own loss. So we are of
42:49
t is equal to r into. I oft vl of t's l into i prime of T and I oft is C into V prime of T. Right. So now what you do is you differentiate the first expression, this expression, you will have this
43:04
Now you have you know V prime of V V C T prime, right? So from here you can
43:10
write it one by C. I of T. So this is straight forward. But for V prime of v r prime You have to derivate this expression. Right? You already derivate this expression. If you derivate, you will have r into i prime of t, so that will be this. now, for V Prime. L. Vls prime t, you derivate this expression? You will have L into I double prime of T. So now, you will have the second derivative with respect to T.
43:44
now if I substitute everything I'll have L into i double prime of T plus R into i prime of T plus 1, by c into i of T is equals to VS of this derivative. Okay, so now we have The second derivative. Therefore this. Equation. So formed is the second order ODA. But the coefficients you see, this is R. Right. So this is R and this is 1 by C. Both are constants. Okay, so all the coefficients even L.
44:18
So the coefficients are this coefficients of this, i double prime of t. I of T and i I prime of T's. They are the coefficients on the right hand side, it can be constant or it can be any other function, okay. so this is second order linear With the constant coefficients. alright, so now if I have to talk in general, the general form of the second order linearity with constant coefficients, will be like this. So, you'll have some constant multiplied to F double prime of X, plus some constant
44:55
multiplative. Prime of X, some constant multiplier. F of X is equal to some function of x. Okay, so ABC are just the real constraints. Okay, so for solution. We have to do some guessworks.
45:11
okay, so we will discuss those things in the next session, but we will, for the
45:18
solution, you not only mathematic completely on analytical approach. We should also do some guesses. Okay, so let me talk the easiest case. the easiest case, or the homogeneous case, Is when the right hand side is 0. so, when you have the right hand side to be 0, Our second order OD will just become A into F double prime of X plus B into F prime of x plus c into F of x is equal to 0. Right. So now what we will do is we'll guess we'll guess that a f of x is Some e
45:56
power sx. Okay, is this a guess? okay, so, whereas is a construct okay, we don't know what s is, but we say The solution will be just an exponential. Function. Okay, so once if you have that, I will just substitute. Okay, so once F of X is E power, E power sx. I'll have F prime of X which is S into E power S x. I will also find F double prime of X. Which is s square into e power, s x. So you just have to derivate f prime of x once again. And I will take all these things. This one, this one
46:36
and this one and substitute them in this equation. If I just substitute them in this equation, I will get this right. So a into e power s x double derivative plus b into e power s x derivative plus c into E bar sx if I just substitute them and pull out, everything has a power Sx I will just pull out. E power sx, common
47:03
what I will have is as square plus b s plus C into E power, s x is equal to 0 and we all know that e power S x will never be 0 because it is an exponential function. We just have to. Make this particular expression to be 0. So that means we just need S. Touch that. As square plus b s plus c being 0. Then e power, s x will be a solution to this particular. Right? So that is what it is, right? So I just substituted them in this particular expression and I got this particular thing. So if
47:38
this becomes 0, then left hand side will be equal to the right hand side. And I will have the solution. For, as we will solve this quadratic equation. Okay. and here also, there are some cases okay, so we ultimately in order to get this solution Solution for the second order. We were boiled down to a quadratic equation. See. We boil down to quadratic equation and solving that quadratic equation will give the values for s and we will substitute those values here.
48:22
And that will be the solution, right? So, so for solving second order already is we are using the idea of quadratic equations and its solutions. Okay. so here again, once we reach to quadratic equation, The solutions can be of three possibilities, right? So There can never be any real number at all.
48:45
Right? So, the equation doesn't have any real numbers that satisfy. Or. The roots. Can be. Distinct. Or the roots can be the same. So even though it's quadratic equation,
49:03
We can have the same roots. Repeated roots, right? So three possible cases. Three possibilities. One is no real roots at all.
49:13
So these things we have learned, okay? You just have to recall Or there can be distinct. Roots. Or they can be one real root repeated.
49:28
Right. So Three possibilities and we will discuss all three possibilities separately. So, the first case we'll deal is Distinct. Real roots. Okay, so with this, we will stop for today and we will discuss the other things in the next session. Okay. So in homogeneous case homogeneous, remember homogeneous means the right hand side is zero.
49:56
And distinct real roots case. So where once you form the quadratic equation you will get two roots, say S1 and S2 Okay. so, once you solve this, you will have S1 and S2 and both are real and though when I say distinct, this is important as one should not be equal to S2.
50:17
Okay, so now the solutions can be E power S, 1 X and E power. S 2 X also. So both are possibilities, right? So once you have these two, Any linear combination of these two will also be a solution. So that's the theorem and I think you can also prove this. I'm not sure whether Professor has proved this or not, if not proved, we will do it on Sunday, okay? But the thing is,
50:43
if you have this as a solution, if you have this as a solution to your ODA, then any linear combination of these two will also be a solution to your Audi. okay, so what it means is
50:58
Any c, 1 times, e power, s, 1 x plus c, 2 times C. Power s 2 x will be a solution for this. So this is the general solution will look like this. so for second order already, what we have to do, homogeneous case what we have
51:14
to do, we have to form this auxiliary equation. Okay. So this is called auxiliary equation. Will form this auxiliary equation. By substituting.
51:26
E Power s x as the solution by guessing the solution as e power, s, x and substituting them in this particular expression. in particular equation, and we will get this After getting to this, we will solve for the roots. Once, if you have the distinct roots,
51:44
Then we know that e power S 1, x e, power, S 2 x are the solutions and any linear combination will also be solution. So we will write the general solution to be like this. After getting roots as S1, and S2 will write f of X to be. C. 1 E Power S 1 X plus C, 2, E power, S 2 x. Where c 1 and c 2 are unknown to us. Okay, C1 and C 2 are still unknown. We don't know. But we will find them using the initial conditions, okay? So initial conditions will be given to us using those initial conditions.
52:14
We will find C1 and C2 and substitute them here and finally write down the expression. Okay, so distinct real roots case. Is very, very easy. You just have to follow to two step. It's a two-step process. so, for example,
52:32
You have this, it is given like this. So f double prime of x minus 6 into F prime of x. Plus 8, into f of x is equal to 0 and the initial conditions are f of 0 is 1 and f prime of 0 is also Okay, so what you have to do first, you have to form the auxiliary equation. If you write down the auxiliary equation, you will see that. If you have guessed it, you will have
52:57
A Into. If you have f double prime of X, you will just have S square. If you have F prime, you will have S. If you have F of X, it will not have anything so a into S square plus B into S plus C into Constant 1. okay, so the same thing here also
53:18
So a here is 1, so 1 into S square, minus 6 into s plus 8. So that will be equal to 0. That will be our quadratic equation which is this S square. Minus 6x, plus 8 is equal to 0. So once you solve this quadratic equation, you will see that. S is equals to. 2 and 4. once you have S is equal to 2 and 4, You write on the general solution as f of x is equal to
53:45
C 1 into E power, 2, X plus C 2, into E power 4, x. Right. How do I get C1 and C2? I will use this initial condition so I will substitute 0 here.
53:59
I'll have f of 0 is equal to c 1, plus c 2, and f of 0 is 1. So c 1. Plus c 2 is equal to 1. How do I use the second condition? I will derivate this. if I derivate this, I will have f prime of x to be
54:16
2, c, 1 into E, power 2 x plus 4 C to into E power 4, x. Now, I will substitute Substitute. X is equal to 0 in this, if you substitute x is equal to 0, you will have F prime of 0 to be 2, C 1 plus 4 c 2. Right. So 2 c 1, plus 4 C. 2 is F prime of 0 and F prime of 0 is 1 given to us. So now you have c 1. Plus c is equal to 1 2, c 1 plus 4 c. Is equals to 1. What is this? This is this the system of linear equations. No. Right. So once you have the system of linear equations with two
54:50
equations and two variables, we know how to find it.
54:55
You find it and you will see that you will your C2 will be minus half and c 1 will be 3. Bit what you have to do. You just have to substitute them back in this particular expression. So if you substitute back, you'll get F of X is equals to c 1 is 3 by 2 into E power 2 x. Minus half into E, power 4, x. That will be our solution. so, one suggestion, Which is like validation. So if you, if you have time in the exam, okay? And you want to validate yourself that whether this solution that
55:28
you got is correct or wrong. What you can do is you just take the solution. Find f prime of x. Find f, double prime of x. And substitute them back in this equation. okay, in this equation and see whether your equation is getting satisfied or not, Substitute and validate. Okay. trust me, you can
56:01
always go wrong while you compute C1 and C2, or while you get this.
56:07
okay, while solving quadratic equations, or while doing this system of solving system of linear equations, we may go wrong. I'm not saying you'll definitely go wrong. We may go wrong. It is always better to validate your solution that you get. Okay. So with that,
56:25
We'll stop here. We'll see what will happen when we have a repeated real roots? we'll have an example. Also, And what if they don't have? If it is not a real root.
56:38
Which are I will get to see that it will have complex real roots. What will happen to that?
56:47
Okay, and Non-homogeneous you can kind of skip but it is good to know. You can go through this. But with respect to examination, point of view will not be giving you non homogeneous or something because it will it will take too much of time. To solve. Okay. But at least you should know. The homogeneous cases. With respect to second order. Alright, so that's all for today. We will meet on Sunday. Thank you.
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