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Hovedindhold

# Del 2 af transformationen af sin(at)

## Video udskrift

Welcome back. We were in the midst of figuring out the Laplace transform of sine of at when I was running out of time. This was the definition of the Laplace transform of sine of at. I said that also equals y. This is going to be useful for us, since we're going to be doing integration by parts twice. So I did integration by parts once, then I did integration by parts twice. I said, you know, don't worry about the boundaries of the integral right now. Let's just worry about the indefinite integral. And then after we solve for y-- let's just say y is the indefinite version of this-- then we can evaluate the boundaries. And we got to this point, and we made the realization, after doing two integration by parts and being very careful not to hopefully make any careless mistakes, we realized, wow, this is our original y. If I put the boundaries here, that's the same thing as the Laplace transform of sine of at, right? That's our original y. So now-- and I'll switch colors just avoid monotony-- this is equal to, actually, let me just-- this is y. Right? That was our original definition. So let's add a squared over sine squared y to both sides of this. So this is equal to y plus-- I'm just adding this whole term to both sides of this equation-- plus a squared over s squared y is equal to-- so this term is now gone, so it's equal to this stuff. And let's see if we can simplify this. So let's factor out an e to the minus st. Actually, let's factor out a negative e to the minus st. So it's minus e to the minus st, times sine of-- well, let me just write 1 over s, sine of at, minus 1 over s squared, cosine of at. I really hope I haven't made any careless mistakes. And so this, we can add the coefficient. So we get 1 plus a squared, over s squared, times y. But that's the same thing as s squared over s squared, plus a squared over s squared. So it's s squared plus a squared, over s squared, y is equal to minus e to the minus st, times this whole thing, sine of at, minus 1 over s squared, cosine of at. And now, this right here, since we're doing everything with respect to dt, this is just a constant, right? So we can say a constant times the antiderivative is equal to this. This is as good a time as any to evaluate the boundaries. Right? If this had a t here, I would have to somehow get them back on the other side. Because the t's are involved in evaluating the boundaries, since we're doing our definite integral or improper integral. So let's evaluate the boundaries now. And we could've kept them along with us the whole time, right? And just factored out this term right here. But anyway. So let's evaluate this from 0 to infinity. And this should simplify things. So the right-hand side of this equation, when I evaluate it at infinity, what is e to the minus infinity? Well, that is 0. We've established that multiple times. And now it approaches 0 from the negative side, but it's still going to be 0, or it approaches 0. What's sine of infinity? Well, sine just keeps oscillating, between negative 1 and plus 1, and so does cosine. Right? So this is bounded. So this thing is going to overpower these. And if you're curious, you can graph it. This kind of forms an envelope around these oscillations. So the limit, as this approaches infinity, is going to be equal to 0. And that makes sense, right? These are bounded between 0 and negative 1. And this approaches 0 very quickly. So it's 0 times something bounded between 1 and negative 1. Another way to view it is the largest value this could equal is 1 times whatever coefficient's on it, and then this is going to 0. So it's like 0 times 1. Anyway, I don't want to focus too much on that. You can play around with that if you like. Minus this whole thing evaluated at 0. So what's e to the minus 0? Well, e to the minus 0 is 1. Right? That's e to the 0. We have a minus 1, so it becomes plus 1 times-- now, sine of 0 is 0. Minus 1 over s squared, cosine of 0. Let's see. Cosine of 0 is 1, so we have minus 1 over s squared, minus 1 over s squared, times 1. So that is equal to minus 1 over s squared. And I think I made a mistake, because I shouldn't be having a negative number here. So let's backtrack. Maybe this isn't a negative number? Let's see, infinity, right? This whole thing is 0. When when you put 0 here, this becomes a minus 1. Yeah. So either this is a plus or this is a plus. Let's see where I made my mistake. e to the minus st-- oh, I see where my mistake is. Right up here. Where I factored out a minus e to the minus st, right? Fair enough. So that makes this 1 over s, sine of at. But if I factor out a minus e to the minus st, this becomes a plus, right? It was a minus here, but I'm factoring out of a minus e to the minus st. So that's a plus. This is a plus. Boy, I'm glad that was not too difficult to find. So then this becomes a plus. And then this becomes a plus. Thank God. It would have been sad if I wasted two videos and ended up with a careless, negative number. Anyway. So now we have s squared plus a squared, over s squared, times y is equal to this. Multiply both sides times s squared over-- s squared plus a squared. Divide both sides by this, and we get y is equal to 1 over s squared-- And actually, let me make sure that that is right. It's 1 over s squared. y is equal to 1 over s squared, times s squared, over s squared plus a squared. And then these cancel out. And let me make sure that I haven't made another careless mistake. Because I have a feeling I have. Yep. There. I see the careless mistake. And it was all in this term. And I hope you don't mind my careless mistakes, but I want you to see that I'm doing these things in real time and I am human, in case you haven't realized already. Anyway, so I made the same careless mistake. So I factor out an e to the minus st here, so it's plus. But it was a over s squared. So this is an a. That's an a. And so this is an a. And so this is an a. And so this is an a. Right? This was an a. And this is the correct answer. a over s squared plus a squared. So I hope those careless mistakes didn't throw you off too much. These things happen when you do integration by parts twice with a bunch of variables. But anyway, now we are ready to add a significant entry into our table of Laplace transforms. And that is that the Laplace transform-- I had an extra curl, there. That was unecessary. Let me do it again. The Laplace transform of sine of at is equal to a over s squared, plus a squared. And that's a significant entry. And maybe a good exercise for you, just to see how fun it is to do these integration by parts problems twice, is to figure out the Laplace transform of cosine of at. And I'll give you a hint. It's s over s squared over s squared plus a squared. And it's nice that there's that symmetry there. Anyway, I'm almost at my time limit. And I'm very tired working on this video. So I'll leave it there and I'll see you in the next one.