"Shifting" transform by multiplying function by exponential

Now I think is a good time to add some notation and techniques to our Laplace Transform tool kit. So the first thing I want to introduce is just kind of a quick way of doing something. And that is, if I had the Laplace Transform, let's say I want to take the Laplace Transform of the second derivative of y. Well, we proved several videos ago that if I wanted to take the Laplace Transform of the first derivative of y, that is equal to s times the Laplace Transform of y minus y of 0. And we used this property in the last couple of videos to actually figure out the Laplace Transform of the second derivative. Because if you just, you know, if you say this is y prime, this is the anti-derivative of it, then you could just pattern match. You could say, well, the Laplace Transform of y prime prime, that's just equal to s times the Laplace Transform of y prime minus y prime of 0. This is the derivative of this, just like this is the derivative of this. I'll draw a line here just so you don't get confused. So the Laplace Transform of y prime prime is this thing. And now we can use this, which we proved several videos ago, to resubstitute it and get it in terms of the Laplace Transform of y. So we can expand this part. The Laplace Transform of the derivative of y, that's just equal to s times the Laplace Transform of y minus y of 0. And then we have the outside, right? We have s minus y prime of 0. And then when you expand it all out, and we've done this before, you get s squared times the Laplace Transform of y minus s times y of 0 minus y prime of 0. Now there's something interesting to note here, and if you learn this it'll make it a lot faster. You won't have to go through all of this and risk making careless mistakes when you have scarce time and paper on your test. Just notice that when you take the Laplace Transform of the second derivative, what do we end up? We end up with s squared, right? This was the second derivative. So I end up with s squared times the Laplace Transform of y, minus s times y of 0 minus 1 times y prime of 0. So every term, we started with s squared, and then every term we lower the degree of s one, and then everything except the first term is a negative sign. And then we started with the Laplace Transform of y, and then you can almost view the Laplace Transform as a kind of integral, so we kind of take the derivatives, so then you get y. And then you take the derivative again, you get y prime. And of course every other term is negative. And these aren't the actual functions. These are those functions evaluated at 0. But that's a good way to help you, hopefully, remember how to do these. And once you get the hang of it, you can take the Laplace Transform of any arbitrary function very, very quickly. Or any arbitrary derivatives. So let's say we wanted to take the Laplace Transform of, I don't know, this should hit the point home, the fourth derivative of y. That 4 in parentheses means the fourth derivative. I could have drawn four prime marks, but either way. So what is this equal to? If we use this technique and substitute it, we're bound to make some form of careless mistake or other, and it would take us forever and it would waste a lot of paper. But now we see the pattern, and so we can just say, well, the Laplace Transform of this, in terms of the Laplace Transform of y, right, that's what we want to get to, is going to be s to the fourth times the Laplace Transform of y-- now every other term is going to have a minus in front of it-- minus-- lower the degree on the s-- minus s to the third. And then you could kind of say, let's take the, you know, so form of derivative, so you get y of 0-- it's not a real derivative. The Laplace Transform really isn't the anti-derivative of y of 0, but anyway, I think you get the idea. And then we lower the degree on s again, minus s squared, take the derivative. And of course these aren't functions. But we're evaluating the derivative of that function now of 0. So y prime of 0, minus-- now we lower the degree one more-- minus s, times-- this is an s-- times y prime prime of 0. We have one more term. Lower the degree on the s one more time. Then you get s to the 0, which is just 1. So minus-- and 1 is a coefficient-- and then you have y, the third derivative of y-- let me scroll over a little bit-- the third derivative of y evaluated at 0. So I think you see the pattern now. And this is a much faster way of evaluating the Laplace Transform of an arbitrary derivative of y, as opposed to keep going through that pattern over and over again. Another thing I want to introduce you to is just a notational savings. And it's just something that you'll see, so you might as well get used to it. And it actually saves time over, you know, keep writing this curly L in this bracket. If I have the Laplace Transform of y of t, I can write as, and people tend to write it as-- well, it's going to be a function of s, and what they use is a capital Y to denote the function of s. It makes sense, because normally when we're doing antiderivatives, you just take-- you know, when you learn the fundamental theorem of calculus, you learn that the integral of f with respect to dx, you know, from 0 to x, is equal to capital F of x. So it's kind of borrowing that notation, because this function of s is kind of an integral of y of t. The Laplace Transform, to some degree, is like a special type of integral where you have a little exponential function in there to mess around with things a little bit. Anyway, I just wanted you to get used to this notation. When you see capital Y of s, that's the same thing as a Laplace Transform of y of t. And you might also see it this way. The Laplace Transform of f of t is equal to capital F of s. And the clue that tells you that this isn't just a normal antiderivative, is the fact that they're using that s as the independent variable. Because in general, s represents the frequency domain, and if people were to use s with just a general antiderivative, people would get confused, et cetera, et cetera. Anyway, I'm trying to think whether I have time to teach you more fascinating concepts of Laplace Transform. Well, sure, I think we do. So my next question for you-- and now we'll teach you a couple more properties, and this'll be helpful in taking Laplace Transforms. What is the Laplace Transform of e to the at times f of t? Fascinating. Well, let's just should go back to our definition of the Laplace Transform. It is the integral from 0 to infinity of e to the minus st times whatever we have between the curly brackets. So, with the curly brackets we have e to the at f of t dt. And now we can add these exponents. We have a similar base, so this is equal to what? This is equal to the integral from 0 to infinity. And let's see, I want to write it as, I could write it minus s plus a, but I'm going to write it as minus s minus a t. And you could expand this out. It becomes minus s plus a, which is exactly what we have here, times f of t dt. Now let me show you something. if I were to just take the Laplace Transform of f of t, that is equal to some function of s. Whatever we essentially have right here for s, it becomes some function of that. So this is interesting. This is some function of s. Here, all we did to go from-- well actually let me rewrite this. The Laplace, which is equal to 0 to infinity e to the minus st f of t dt. The Laplace Transform of just f of t is equal to this, which is some function of s. Well, the Laplace Transform of e to the at, times f of t, it equals this. And what's the difference between this and this? What's the difference between the two? Well, it's not much. Here, wherever I have an s, I have an s minus a here. So if this is a function of s, what's this going to be? It's going to be that same function. Whatever the Laplace Transform of f was, it's going to be that same function, but instead of s, it's going to be a function of s minus a. And once again, how did I get that? Well I said the Laplace Transform of f is a function of s, and it's equal to this. Well if I just replace an s with an s minus a, I get this, which is a function of s minus a. Which was the Laplace Transform of e to the at times f of t. Maybe that's a little confusing. Let me show you an example. Let's just take the Laplace Transform of cosine of 2t. We've shown is equal to-- well I'll write the notation-- it's equal to some function of s. And that function of s is s over s squared plus 4. We've shown that already. And so the Laplace Transform of e to the, I don't know, 3t times cosine of 2t, is going to be equal to the same function, but instead of s, it's going to be a function of s minus a. So s minus 3, which is equal to s minus 3 over s minus 3 squared plus 4. Notice, when you just multiply something by this, either the 3t and then or either the at, you take the Laplace Transform of it, you just-- it's the same thing as the Laplace Transform of this function, but everywhere where you had an s, you replace it with an s minus this a. Anyway, I hope I didn't confuse you too much with that last part. I think my power adaptor actually just went on. I hope the video keeps recording. I'll see you in the next one.