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## Differential equations

### Course: Differential equations>Unit 3

Lesson 3: Laplace transform to solve a differential equation

# Laplace transform to solve an equation

Using the Laplace Transform to solve an equation we already knew how to solve. Created by Sal Khan.

## Want to join the conversation?

• Is there a known good source for learning about Fourier transforms, which Sal mentions in the beginning? I can't find it on Khan Academy.
(24 votes)
• I made the video I always wanted to watch about Fourier series, I am going to put the link to the first part here

https://www.youtube.com/watch?v=72eXzIjou2g&feature=youtu.be
(11 votes)
• Hey Sal, at you recommend to learn about Fourier series and Fourier Transforms. I tried to find videos about this but I didn't find any.
(15 votes)
• I found this on Youtube, it's pretty decent: https://www.youtube.com/watch?v=x04dnqg-iPw
(5 votes)
• Can you please do videos on Fourier series?
(10 votes)
• Hello.

If it is ty "(t), we apply twice the formula of the derivative (to obtain the TL of y"), then the formula of multiplication by t. As you might have thought yourself.

Good work !
(1 vote)
• Sir can you please do video on the heavy-side function as well please
(5 votes)
• *heaviside functions named after Oliver Heaviside. Pronunciation is the same though.
(4 votes)
• what is fourier transform
(2 votes)
• Sir,in the beginning of the problem you've applied the laplace transformation operator on both sides are there any precautions for that step, like you know, any necessary conditions or something
(2 votes)
• There is an axiom known as the axiom of substitution which says the following: if `x` and `y` are objects such that `x = y`, then we have `ƒ(x) = ƒ(y)` for every function `ƒ`. Hence, when we apply the Laplace transform to the left-hand side, which is equal to the right-hand side, we still have equality when we also apply the Laplace transform to the right-hand side by the axiom of substitution. Is this what you had in mind?
(3 votes)
• What if we don't know the values of y(0) and y'(0) and instead know other values of y and y', for example, y(1) or y'(1) or something alike. How would we solve the math ?
(3 votes)
• Your videos have helped me to understand transforms much better than when my D.E. teacher was explaining them, and I've really come to appreciate them. So my question is, can I use Laplace transforms for all of these types of equations, or is it better to look at each problem as a case by case basis, and determine which method to use from there? Sorry if this sounds confusing...
(3 votes)
• why is it that L(0) = integral 0 between 0 and infinity equal to zero
but the indefinite integral of 0 equal to a constant?
(1 vote)
• No, he's not saying the integral of 0 = C.
That constant comes from the other side of the equation. That integral = something + C and then he subtracts C from both sides.

Hope I understood your question:)
(3 votes)
• Sal says that L{y'}=sL{y}-y(0), where do we get that? He also says that we already proved it. Am I missing a video that he uploaded or something?
(1 vote)

## Video transcript

I've been doing a ton of videos on the mechanics of taking the Laplace Transform, but you've been sitting through them always wondering, what am I learning this for? And now I'll show you, at least in the context of differential equations. And I've gotten a bunch of letters on the Laplace Transform. What does it really mean? And and all that. And those are excellent questions and you should strive for that. It's hard to really have an intuition of the Laplace Transform in the differential equations context, other than it being a very useful tool that converts differential or integral problems into algebra problems. But I'll give you a hint, and if you want a path to learn it in, you should learn about Fourier series and Fourier Transforms, which are very similar to Laplace Transforms. And that'll actually build up the intuition on what the frequency domain is all about. Well anyway, let's actually use the Laplace Transform to solve a differential equation. And this is one we've seen before. So let me see. So let's say the differential equation is y prime prime, plus 5, times the first derivative, plus 6y, is equal to 0. And you know how to solve this one, but I just want to show you, with a fairly straightforward differential equation, that you could solve it with the Laplace Transform. And actually, you end up having a characteristic equation. And the initial conditions are y of 0 is equal to 2, and y prime of 0 is equal to 3. Now, to use the Laplace Transform here, we essentially just take the Laplace Transform of both sides of this equation. Let me use a more vibrant color. So we get the Laplace Transform of y the second derivative, plus-- well we could say the Laplace Transform of 5 times y prime, but that's the same thing as 5 times the Laplace Transform-- y prime. y prime plus 6 times the Laplace Transform of y. And let me ask you a question. What's the Laplace Transform of 0? Let me do that. So the Laplace Transform of 0 would be be the integral from 0 to infinity, of 0 times e to the minus stdt. So this is a 0 in here. So this is equal to 0. So the Laplace Transform of 0 is 0. And that's good, because I didn't have space to do another curly L. So what are the Laplace Transforms of these things? Well this is where we break out one of the useful properties that we learned. Let me write it over here. I think that's going to need as much real estate as possible. Let me erase this. So we learned that the Laplace Transform-- I'll do it here. Actually, I'll do it down here. The Laplace Transform of f prime, or we could even say y prime, is equal to s times the Laplace Transform of y, minus y of 0. We proved that to you. And this is extremely important to know. So let's see if we can apply that. So the Laplace Transform of y prime prime, if we apply that, that's equal to s times the Laplace Transform of-- well if we go from y prime to y, you're just taking the anti-derivative, so if you're taking the anti-derivative of y, of the second derivative, we just end up with the first derivative-- minus the first derivative at 0. Notice, we're already using our initial conditions. I won't substitute it just yet. And then we end up with plus 5, times-- I'll write it every time-- so plus 5 times the Laplace Transform of y prime, plus 6 times the Laplace Transform of y. All of that is equal to 0. So just to be clear, all I did is I expanded this into this using this. So how can we rewrite the Laplace Transform of y prime? Well, we could use this once again, so let's do that. So this over here-- I'll do it in magenta-- this is equal to s times what? s times the Laplace Transform of y prime. Well that's s times the Laplace Transform of y, minus y of 0, right? I took this part and replaced it with what I have in parentheses. So minus y prime of 0-- and now I'll switch colors-- plus 5 times-- once again the Laplace Transform of y prime. Well we can use this again. So 5 times s times Laplace Transform of y, minus y of 0, plus 6 times the Laplace Transform-- oh I ran out of space, I'll do it in another line-- plus 6 times the Laplace Transform of y. All of that is equal to 0. I know this looks really confusing but we'll simplify right now. And we could get rid of this right here, because we've used it as much as we need to. So now we just simplify. And notice, using the Laplace Transform, we didn't have to guess at a general solution or anything like that. Even when we did a characteristic equation, we guessed what the original general solution was. Now we're just taking Laplace Transforms, and let's see where this gets us. And actually I just want to make clear, because I know it's very confusing, so I rewrote this part as this. And I rewrote this thing as this. And everything else is the same. But now let's simplify the math. So we get s squared, times the Laplace Transform of y-- I'm going to write smaller, I've learned my lesson-- minus s times y of 0. Let's substitute y of 0 here. y of 0 is 2, so s times y of 0 is 2 times s, so 2s, distribute that s, minus y prime of 0. Y prime of 0 is 3. So minus 3, plus-- so we have 5 times s times the Laplace Transform of y, so plus 5s times the Laplace Transform of y, minus 5 times y of 0. y of 0 is 2, so minus 10. Minus 10, right? 5 times-- this is 2 right here-- so 5 times 2, plus 6 times the Laplace Transform of y. All of that is equal to 0. Now, let's group our Laplace Transform of y terms and our constant terms, and we should be hopefully getting some place. So let's see, my Laplace Transform of y terms, I have this one, I have this one, and I have that one. So what am I left with? Well let me factor out the Laplace Transform of y part. So I get the Laplace Transform of y-- and that's good because it's a pain to keep writing it over and over-- times s squared plus 5s plus 6. So those are all my Laplace Transform terms. And then I have my constant terms. So let's see, I have 1s, so minus 2s, minus 3, minus 10, is equal to 0. And what can we do here? Well, this is interesting, first of all. Notice that the coefficients on the Laplace Transform of y terms, that those are that characteristic equation that we dealt with so much, and that is hopefully, to some degree, second nature to you. So that's a little bit of a clue, and if you want some very tenuous connections, well that makes a lot of sense. Because the characteristic equation to get that, we substituted e to the rt, and the Laplace Transform involves very similar function. But anyway, let's go back to the problem. So how do we solve this? And actually, let me just give you the big picture here, because this is a good point. What I'm going to do is I'm going to solve this. I'm going to say the Laplace Transform of y is equal to something. And then I'm going to say, boy, what functions the Laplace Transform is at something? And then I'll have the solution. If that confuses you, just wait and hopefully it'll make some sense. From here until that point it's just some fairly hairy algebra. So let's scroll down a little bit, just so we have some breathing room. And so I get the Laplace Transform of y, times s squared, plus 5s, plus 6, is equal to-- let's add these terms to both sides of this equation-- is equal to 2s plus 3 plus 10-- oh, that's silly-- plus 13. This is minus 13 here. A phone call. Who's calling? I think it's some kind of marketing phone call. Anyway, 2s plus 13, and now what can I do? Well. Let's divide both sides by this s squared plus 5s plus 6. So I get the Laplace Transform of y is equal to 2s plus 13, over s squared plus 5s plus 6. Now we're almost done. Everything here is just a little bit of algebra. So now we're almost done. We haven't solved for y yet, but we know that the Laplace Transform of y is equal to this. Now, if we just had this in our table of our Laplace Transforms, we would immediately know what y was, but I don't see something, or I don't remember anything we did in our table that looks like this expression of s. I'm essentially out of time, so the next video we're going to figure out what functions Laplace Transform is this. And it actually turns out it's a sum of things we already know, and we just have to manipulate this a little bit algebraically. See you in the next video.