# Brug af Laplace-transformation til at løse en ikke-homogen ligning

## Video udskrift

It's been over a year since I last did a video with the differential equations playlist, and I thought I would start kicking up, making a couple of videos. And I think where I left, I said that I would do a non-homogenous linear equation using the Laplace Transform. So let's do one, as a bit of a warm-up, now that we've had a-- or at least I've had a one-year hiatus. Maybe you're watching these continuously, so you're probably more warmed up than I am. So if we have the equation the second derivative of y plus y is equal to sine of 2t. And we're given some initial conditions here. The initial conditions are y of 0 is equal to 2, and y prime of 0 is equal to 1. And where we left off-- and now you probably remember this. You probably recently watched the last video. To solve these, we just take the Laplace Transforms of all the sides. We solve for the Laplace Transform of the function. Then we take the inverse Laplace Transform. If that doesn't make sense, then let's just do it in this video, and hopefully the example will clarify all confusion. So in the last video-- it was either the last one or the previous one-- I showed you that the Laplace Transform of the second derivative of y is equal to s squared times the Laplace Transform of y-- and we keep lowering the degree on s-- so minus s times y of 0. You can kind of think of it as taking the derivative. This is an integral. It's not exactly the anti-derivative of this. But the Laplace Transform it is an integral. The Transform is an integral. So y of 0 is kind of one derivative away from that. And then minus y prime of 0. And then we could also rewrite this. And this is just a purely notational issue. I could write this, instead of writing the Laplace Transform of y all the time, I could write this as s squared times capital Y of s-- because this is going to be a function of s, not a function of y-- minus s times y of 0 minus y prime y of 0. These are going to be numbers, right? These aren't functions. These are the function evaluated at 0, or the derivative of the function evaluated at 0. And we know what these values are. y of 0, right here, is 2, and y prime of 0 is 1. It was given to us. So if we take the Laplace Transforms of both sides of this equation, first we're going to want to take the Laplace Transform of this term right there, which we've really just done. The Laplace Transform of the second derivative is s squared times the Laplace Transform of the function, which we write as capital Y of s, minus this, minus 2s-- they gave us that initial condition-- and then minus 1. Right? This term right here is just 1, so minus 1. So that's term right there. Then we want to take the Laplace Transform of y by itself. So this is just plus Y of s, right, the Laplace Transform of y. So I'll just rewrite Laplace Transform of y. I'm just rewriting it in this notation. Y of S. It's good to get used to either one. This is going to be equal to the Laplace Transform of sine of 2t. And I showed you in a video last year that we showed what the Laplace Transform of sine of at is, but I'll write it down here just so you remember it. Laplace Transform of the sine of at is equal to a over s squared plus a squared. So the Laplace Transform of sine of 2t. Here, a is 2. This is going to be 2 over s squared plus 4. So if we take the Laplace Transform of both sides of this, the right-hand side is going to be 2 over s squared plus 4. Now what we can do is we can separate out all the Y of s terms. And so we can factor, well I guess we could say, factor out their coefficients, so that's a Y of s term, that's a Y of s term. And so we could write the left-hand side here as s squared-- that's that term-- plus 1-- the coefficient on that term-- s squared plus 1, times Y of s. Let me do it in green. So this is Y of s, and this Y of s, times Y of s, and then we have the non-Y of s terms. These two right here. So minus 2s, minus 1, is equal to 2 over s squared plus 4. We can add 2s plus 1 to both sides, to essentially move this to the right-hand side, and we're left with s squared plus 1, times Y of s, is equal to 2 over s squared plus 4, plus 2s, plus 1. Now we can divide both sides of this equation by s squared plus 1, and we get the Laplace Transform of Y. Y of s is equal to-- let me switch colors-- it's equal to 2 over s squared plus 4 times this thing right here. I'm dividing both sides of this equation by this term right there. So times s squared plus 1-- it's in the denominator so I'm dividing by it-- plus 2s plus 1-- I have to divide both of those terms by the s squared plus 1-- divided by s squared plus 1, divided by s squared plus 1. Now, in order to be able to take the inverse Laplace Transform of this, I need to get it in some type of simple fraction form. These are actually easier to do, but this was one's a little bit difficult. I want to do some partial fraction decomposition to break this up into maybe simpler fractions. So what I want to do, I'm going to do a little bit of an aside here. And this really is the hardest part of these problems. The algebra, breaking this thing up. So I'm going to break this up. So let me write this this way. 2 over s squared plus 4 times s squared plus 1. I'm going to break this up into two fractions. This is the partial fraction decomposition. One fraction is s squared plus 4. And the other fraction is s squared plus 1. And since both of these denominators are of degree 2, the numerators are going to be of degree 1. So they're going to be some-- let me write it this way-- this one will be As plus B. And then this one will be Cs plus D. This is just pure algebra. This is just partial fraction decomposition. I've made a couple of videos on it. And I'm saying that I'm assuming that this expression right here can be broken up into two expressions of this form. And I need to solve for A, B, C, and D. So let's see how we can do that. So if I were to start with these two and add them up, what do I get? I would have to multiply these times-- so my denominator, my common denominator, would be this thing again-- it would be s squared plus 4 times s squared plus 1. And now I'm going to have to multiply the As plus B times this s squared plus 1. This, as it is right now, these two terms would cancel out. You'll just get this term, but I need to add it to this right here. So you get plus Cs plus D times s squared plus 4. And now let's see what we could do to match up the terms here with this number 2 right here. So let's multiply all of this out. So As times s squared is As to the 3rd. As times 1 is plus As. B times s squared, so plus Bs squared, and then you have B times 1 is plus B. And then you have Cs times s squared, that's Cs to the 3rd. And then Cs times 4, so it's plus 4Cs. These problems are tiring. And I also have a cold, so this is especially tiring, but I'll soldier forward. Where was I? So I multiplied the C's times each of these, now I have to multiply the D's. So plus Ds squared-- that's D times that one-- plus D times 4, so plus 4D. So that's all of them. And I just wrote it this way so I have the common degree terms under each other. So if I were to add the entire numerator, I get-- and I'll just switch colors, somewhat arbitrarily-- I get A plus C times s to the 3rd plus-- let me write the s squared term next-- plus B plus D times s squared-- now I'll write this s term-- plus A plus 4C times s plus B plus 4D. This is just the numerator. This is when I just added these two things up. This whole thing up here simplifies to this. I don't know if the word simplify is appropriate. But it becomes this expression right here. And that's just the numerator. The denominator is still what we had written before. The denominator is still the s squared plus 4, times the s squared plus 1. Of course, I have to show that this is a fraction. And this is going to be equal to this thing over here. 2 over s squared plus 4 times s squared plus 1. Now, why did I go through this whole mess right here? Well, the reason why I went through it is because we should be able to solve for A, B, C, and D. So let's see, A plus C. This is the coefficient on the s cubed term. Do we see any s cubed terms here? No, we see no s cubed terms here. So A plus C-- let me write this down-- A plus C must be equal to 0, because we see nothing here that has an s to the third. B plus D is a coefficient on the s squared term. Do we see any s squared terms here? No, so B plus D must be equal to 0. A plus 4C are the coefficient of the s term. I see no s term over here. So A plus 4C must also be equal to 0. And then finally, we look at just the constant terms. And we do have a constant term on the left-hand side of this equation. We have 2. so B plus 4D-- I didn't want to make it that thick-- B plus 4D must be equal to 2. This just seems like these linear equations are pretty easy to solve for. Let's subtract this from this. So A-- or let me subtract the bottom one from the top one-- so A minus A, that's 0A. And then C minus 4C minus 3C is equal to 0. And so you get C is equal to 0. If C is equal to 0, A plus C is equals to 0, A must be equal to 0. And let's do the same thing here. Let's subtract this from that. So you get B minus B is 0, and then minus 3D-- that's just D minus 4D-- and then 0 minus 2 is equal to minus 2. And then you get D is equal to-- what do we get?-- D is equal to 2/3. Minus 2 divided by minus 3 is 2/3, and then-- this isn't a minus here, I wrote that there later-- we said B plus D is equal to 0. So B must be the opposite of D, right? We could write B is equal to minus D, or B is equal to minus 2/3. Let's remember all of this and go back to our original problem, because we've kind of-- actually let me just be clear. We can rewrite 2 over s squared plus 4 times s squared plus 1. We can rewrite this as, well, A is 0, B is minus 2/3. So this is equal to minus 2/3 over s squared plus 4. And then C is 0, we figured that out. And then D is 2/3. So plus 2/3 over s squared plus 1. So all of that work that I just did, that was just to break up this piece right here. That was just to break up that piece right there. And then, of course, we have these other two pieces here that we can't forget about. So after all of this work, what do we have? And I'm going to make sure I don't make a careless mistake here. We get the Laplace Transform of Y-- as you can see, the algebra is the hardest part here-- is equal to this first term-- I'm just going back-- this first term, which I've now decomposed into this. So it's minus-- let me write it this way-- minus 1/3-- and I think you're going to see in a second why I'm writing this way-- minus 1/3 times 2 over s squared plus 4, and then plus 2/3 times 1 over s squared plus 1. And you're probably saying, Sal, why are you writing it this way? Well you can already immediately see that this is the Laplace Transform of sine of 2t. This is the Laplace Transform of sine of t. So I wanted to write this 2 here, because this is 2, this is 2 squared. This is 1, this is 1 squared. So I wanted to write it in this form. This was just the first term. We have two more terms to worry about. I don't want to make a careless mistake. I have 2s over s squared plus 1. So let me write that down. So plus 2 times s over s squared plus 1, plus-- last one-- plus 1 over s squared plus 1. Now we just take the inverse Laplace Transform of the whole thing, and then we'll know what Y is. So, you know, just to remember the Laplace Transform. So this is going to be a little inverse. This is going to be sine of 2t. Let me just write, just so we have it here, so you know I'm not doing some type of voodoo. The Laplace Transform of sine of at is equal to a over s squared plus a squared. And the Laplace Transform of cosine of at is equal to s over s squared plus a squared. Let's just remember those two things when we take the inverse Laplace Transform of both sides of this equation. The inverse Laplace Transform of the Laplace Transform of y, well that's just y. y-- maybe I'll write it as a function of t-- is equal to-- well this is the Laplace Transform of sine of 2t. You can just do some pattern matching right here. If a is equal to 2, then this would be the Laplace Transform of sine of 2t. So it's minus 1/3 times sine of 2t plus 2/3 times-- this is the Laplace Transform of sine of t. If you just make a is equal to 1, sine of t's Laplace Transform is 1 over s squared plus 1. So plus 2/3 times the sine of t-- let me do the next one in blue, just because it was already written in blue-- plus 2 times-- this is the Laplace Transform of cosine of t. If you make a is equal to 1, then the cosine t Laplace Transform is s over s squared plus 1. So 2 times cosine of t-- and then one last term-- plus-- this is just like this one over here, this is just the Laplace Transform of sine of t-- plus sine of t. And we're almost done. We're essentially done, but there's a little bit more simplification we can do. I have 2/3 times the sign of t here, and then I have another 1 sine of t here, so I can add the 2/3 to the 1. What's 2/3 plus 1, or 3/3? It's 5/3. So I can write y of t is equal to minus 1/3 sine of 2t plus-- these two terms I'm just going to add up-- plus 5/3 sine of t. And then I have this last term here, plus 2cosine of t. So this was a hairy problem, a lot of work. And we saw that the hardest part really was just the partial fraction decomposition that we did up here, not making any careless mistakes. But at the end, we got a pretty neat answer that's not too complicated, that satisfies this non-homogenous differential equation. We were able to incorporate the boundary conditions as we did it. Anyway, hopefully you found that vaguely satisfying. This is a good warm-up after a year of no differential equations.