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# Laplace transform solves an equation 2

## Video transcript

Welcome back. We're finally using the Laplace Transform to do something useful. In the first part of this problem, we just had this fairly straightforward differential equation. And I know it's a little bit frustrating right now, because you're like, this is such an easy one to solve using the characteristic equation. Why are we doing Laplace Transforms? Well I just want to show you that they can solve even these problems. But later on there are going to be classes of problems that, frankly, our traditional methods aren't as good as the Laplace Transform. But anyway, how did we solve this? We just took the Laplace Transform of both sides of this equation. We got all of this hairy mess. We used the property of the derivative of functions, where you take the Laplace Transform, and we ended up, after doing a lot of algebra essentially, we got this. We got the Laplace Transform of y is equal to this thing. We just took the Laplace Transform of both sides and manipulated algebraically. So now our task in this video is to figure out what y's Laplace Transform is this thing? And essentially what we're trying to do, is we're trying to take the inverse Laplace Transform of both sides of this equation. So another way to say it, we could say that y-- if we take the inverse Laplace Transform of both sides-- we could say that y is equal to the inverse Laplace Transform of this thing. 2s plus 13, over s squared plus 5s plus 6. Now we'll eventually actually learn the formal definition of the inverse Laplace Transform. How do you go from the s domain to the t domain? Or how do you go from the frequency domain to the time domain? We're not going to worry about that right now. What we're going to do is we're going to get this into a form that we recognize, and say, oh, I know those functions. That's the Laplace Transform of whatever and whatever. And then we'll know what y is. So let's try to do that. So what we're going to use is something that you probably haven't used since Algebra two, which is I think when it's taught in, you know, eighth, ninth, or 10th grade, depending. And you finally see it now in differential equations that it actually has some use. Let me write it. We're going to use partial fraction expansion. And I'll do a little primer on that, in case you don't remember it. So anyway, let's just factor the bottom part right here. And you'll see where I'm going with this. So if I factor the bottom, I get s plus 2 times s plus 3. And what we want to do, is we want to rewrite this fraction as the sum of 2-- I guess you could call it partial fractions. I think that's why it's called partial fraction expansion. So we want to write this as a sum of A over s plus 2, plus B over s plus 3. And if we can do this, then-- and bells might already be ringing in your head-- we know that these things that look like this are the Laplace Transform of functions that we've already solved for. And I'll do a little review on that in a second. But anyway, how do we figure out A and B? Well if we were to actually add A and B, if we were to-- let's do a little aside right here-- so if we said that A-- so if we were to give them a common denominator, which is this, s plus 2 times s plus 3. Then what would A become? We'd have to multiply A times s plus 3, right? So we'd get As plus 3A. This, as I've written it right now, is the same thing as A over s plus 2. You could cancel out an s plus 3 in the top and the bottom. And now we're going to add the B to it. So plus-- I'll do that in a different color-- plus-- well, if we have this as the denominator, we could multiply the numerator and the denominator by s plus 2, right? To get B times s, plus 2B, and that's going to equal this thing. And all I did is I added these two fractions. Nothing fancier than there. That was Algebra two. Actually, I think I should do an actual video on that as well. But that's going to equal this thing. 2s plus 13, all of that over s plus 2 times s plus 3. Notice in all differential equations, the hairiest part's always the algebra. So now what we do is we match up. We say, well, let's add the s terms here. And we could say that the numerators have to equal each other, because the denominators are equal. So we have A plus Bs plus 3A plus 2B is equal to 2s plus B. So the coefficient on s, on the right-hand side, is 2. The coefficient on the left-hand side is A plus B, so we know that A plus B is equal to 2. And then on the right-hand side, we see 3A plus 2B must be equal to-- oh, this is a 13. Did I say B? This is a 13. That's a 13. It looks just like a B, right? That was 2s plus 13. Anyway, so on the right-hand side I get, it was 3A plus 2B is equal to 13. Now we have two equations with two unknowns, and what do we get? I know this is very tiresome, but it'll be satisfying in the end. Because you'll actually solve something with the Laplace Transform. So let's multiply the top equation by 2, or let's just say minus 2. So we get minus 2A minus 2B equals minus 4. And then we get-- add the two equations-- you get A is equal to-- these cancel out-- A is equal to 9. Great. If A is equal to 9, what is B equal to? B is equal to 9 plus what is equal to 2? Or 2 minus 9 is minus 7. And we have done some serious simplification. Because now we can rewrite this whole expression as the Laplace Transform of y is equal to A over s plus 2, is equal to 9 over s plus 2, minus 7 over s plus 3. Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2, minus 7 times 1 over s plus 3. Why did I take the trouble to do this? Well hopefully, you'll recognize this was actually the second Laplace Transform we figured out. What was that? I'll write it down here just so you remember it. It was the Laplace Transform of e to the at, was equal to 1 over s minus a. That was the second Laplace Transform we figured out. So this is interesting. This is the Laplace Transform of what? So if we were to take the inverse Laplace Transform-- actually let me just stay consistent. So that means that this is the Laplace Transform of y, is equal to 9 times the Laplace Transform of what? If we just do pattern matching, if this is s minus a, then a is minus 2. So 9 times the Laplace Transform of e to the minus 2t. Does that make sense? Take this, put it in this one, which we figured out, and you get 1 over s plus 2. And let me clean this up a little bit, because I'm going to need that real estate. I'll write this. I'll leave that there, because we'll still use that. And then we have minus 7 times-- this is the Laplace Transform of what? This is the Laplace Transform of e to the minus 3t. This pattern matching, you're like, wow, if you saw this, you would go to your Laplace Transform table, if you didn't remember it, you'd see this. You're like, wow, that looks a lot like that. I just have to figure out what a is. I have s plus 3. I have s minus a. So in this case, a is equal to minus 3. So if a is equal to minus 3, this is the Laplace Transform of e to the minus 3t. So now we can take the inverse Laplace-- actually, before we do that. We know that because the Laplace Transform is a linear operator-- and actually now I can delete this down here-- we know that the Laplace Transform is a linear operator, so we can write this. And you normally wouldn't go through all of these steps. I just really want to make you understand what we're doing. So we could say that this is the same thing as the Laplace Transform of 9e to the minus 2t, minus 7e to the minus 3t. Now we have something interesting. The Laplace Transform of y is equal to the Laplace Transform of this. Well if that's the case, then y must be equal to 9e to the minus 2t, minus 7e to the minus 3t. And I never proved to you, but the Laplace Transform is actually a 1:1 Transformation. That if a function's Laplace Transform, if I take a function against the Laplace Transform, and then if I were take the inverse Laplace Transform, the only function whose Laplace Transform that that is, is that original function. It's not like two different functions can have the same Laplace Transform. Anyway, a couple of things to think about here. Notice, we had that thing that kind of looked like a characteristic equation pop up here and there. And we still have to solve a system of two equations with two unknowns. Those are both things that we had to do when we solve an initial value problem, when we use just traditional, the characteristic equation. But here it happened all at once. And frankly it was a little bit hairier because we had to do all this partial fraction expansion. But it's pretty neat. The Laplace Transform got us something useful. In the next video I'll actually do a non-homogeneous equation, and show you that the Laplace Transform applies equally well there. So it's kind of a more consistent theory of solving differential equations, instead of kind of guessing solutions, and solving for coefficients and all of that. See you in the next video.