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Now I think is a good time to add some notation and

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techniques to our Laplace Transform tool kit.

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So the first thing I want to introduce is just kind of a

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quick way of doing something.

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And that is, if I had the Laplace Transform, let's say I

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want to take the Laplace Transform of the second

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derivative of y.

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Well, we proved several videos ago that if I wanted to take

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the Laplace Transform of the first derivative of y, that is

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equal to s times the Laplace Transform of y minus y of 0.

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And we used this property in the last couple of videos to

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actually figure out the Laplace Transform of the

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second derivative.

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Because if you just, you know, if you say this is y prime,

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this is the anti-derivative of it, then you could just

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pattern match.

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You could say, well, the Laplace Transform of y prime

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prime, that's just equal to s times the Laplace Transform of

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y prime minus y prime of 0.

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This is the derivative of this, just like this is the

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derivative of this.

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I'll draw a line here just so you don't get confused.

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So the Laplace Transform of y prime prime is this thing.

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And now we can use this, which we proved several videos ago,

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to resubstitute it and get it in terms of the Laplace

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Transform of y.

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So we can expand this part.

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The Laplace Transform of the derivative of y, that's just

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equal to s times the Laplace Transform of y minus y of 0.

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And then we have the outside, right?

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We have s minus y prime of 0.

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And then when you expand it all out, and we've done this

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before, you get s squared times the Laplace Transform of

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y minus s times y of 0 minus y prime of 0.

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Now there's something interesting to note here, and

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if you learn this it'll make it a lot faster.

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You won't have to go through all of this and risk making

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careless mistakes when you have scarce time and paper on

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your test. Just notice that when you take the Laplace

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Transform of the second derivative, what do we end up?

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We end up with s squared, right?

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This was the second derivative.

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So I end up with s squared times the Laplace Transform of

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y, minus s times y of 0 minus 1 times y prime of 0.

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So every term, we started with s squared, and then every term

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we lower the degree of s one, and then everything except the

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first term is a negative sign.

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And then we started with the Laplace Transform of y, and

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then you can almost view the Laplace Transform as a kind of

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integral, so we kind of take the derivatives, so

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then you get y.

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And then you take the derivative

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again, you get y prime.

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And of course every other term is negative.

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And these aren't the actual functions.

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These are those functions evaluated at 0.

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But that's a good way to help you, hopefully, remember how

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to do these.

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And once you get the hang of it, you can take the Laplace

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Transform of any arbitrary function very, very quickly.

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Or any arbitrary derivatives.

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So let's say we wanted to take the Laplace Transform of, I

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don't know, this should hit the point home, the fourth

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derivative of y.

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That 4 in parentheses means the fourth derivative.

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I could have drawn four prime marks, but either way.

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So what is this equal to?

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If we use this technique and substitute it, we're bound to

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make some form of careless mistake or other, and it would

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take us forever and it would waste a lot of paper.

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But now we see the pattern, and so we can just say, well,

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the Laplace Transform of this, in terms of the Laplace

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Transform of y, right, that's what we want to get to, is

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going to be s to the fourth times the Laplace Transform of

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y-- now every other term is going to have a minus in front

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of it-- minus-- lower the degree on the s--

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minus s to the third.

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And then you could kind of say, let's take the, you know,

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so form of derivative, so you get y of 0-- it's not a real

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derivative.

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The Laplace Transform really isn't the anti-derivative of y

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of 0, but anyway, I think you get the idea.

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And then we lower the degree on s again, minus s squared,

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take the derivative.

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And of course these aren't functions.

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But we're evaluating the derivative of that

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function now of 0.

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So y prime of 0, minus-- now we lower the degree one more--

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minus s, times-- this is an s-- times y prime prime of 0.

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We have one more term.

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Lower the degree on the s one more time.

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Then you get s to the 0, which is just 1.

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So minus-- and 1 is a coefficient-- and then you

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have y, the third derivative of y-- let me scroll over a

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little bit-- the third derivative of y

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evaluated at 0.

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So I think you see the pattern now.

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And this is a much faster way of evaluating the Laplace

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Transform of an arbitrary derivative of y, as opposed to

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keep going through that pattern over and over again.

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Another thing I want to introduce you to is just a

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notational savings.

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And it's just something that you'll see, so you might as

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well get used to it.

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And it actually saves time over, you know, keep writing

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this curly L in this bracket.

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If I have the Laplace Transform of y of t, I can

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write as, and people tend to write it as-- well, it's going

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to be a function of s, and what they use is a capital Y

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to denote the function of s.

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It makes sense, because normally when we're doing

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antiderivatives, you just take-- you know, when you

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learn the fundamental theorem of calculus, you learn that

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the integral of f with respect to dx, you know, from 0 to x,

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is equal to capital F of x.

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So it's kind of borrowing that notation, because this

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function of s is kind of an integral of y of t.

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The Laplace Transform, to some degree, is like a special type

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of integral where you have a little exponential function in

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there to mess around with things a little bit.

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Anyway, I just wanted you to get used to this notation.

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When you see capital Y of s, that's the same thing as a

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Laplace Transform of y of t.

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And you might also see it this way.

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The Laplace Transform of f of t is equal to capital F of s.

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And the clue that tells you that this isn't just a normal

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antiderivative, is the fact that they're using that s as

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the independent variable.

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Because in general, s represents the frequency

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domain, and if people were to use s with just a general

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antiderivative, people would get confused,

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et cetera, et cetera.

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Anyway, I'm trying to think whether I have time to teach

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you more fascinating concepts of Laplace Transform.

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Well, sure, I think we do.

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So my next question for you-- and now we'll teach you a

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couple more properties, and this'll be helpful in taking

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Laplace Transforms. What is the Laplace Transform of e to

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the at times f of t?

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Fascinating.

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Well, let's just should go back to our definition of the

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Laplace Transform.

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It is the integral from 0 to infinity of e to the minus st

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times whatever we have between the curly brackets.

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So, with the curly brackets we have e to the at f of t dt.

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And now we can add these exponents.

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We have a similar base, so this is equal to what?

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This is equal to the integral from 0 to infinity.

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And let's see, I want to write it as, I could write it minus

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s plus a, but I'm going to write it as minus s minus a t.

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And you could expand this out.

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It becomes minus s plus a, which is exactly what we have

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here, times f of t dt.

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Now let me show you something.

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if I were to just take the Laplace Transform of f of t,

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that is equal to some function of s.

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Whatever we essentially have right here for s, it becomes

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some function of that.

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So this is interesting.

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This is some function of s.

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Here, all we did to go from-- well actually

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let me rewrite this.

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The Laplace, which is equal to 0 to infinity e to the minus

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st f of t dt.

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The Laplace Transform of just f of t is equal to this, which

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is some function of s.

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Well, the Laplace Transform of e to the at, times f of t, it

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equals this.

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And what's the difference between this and this?

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What's the difference between the two?

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Well, it's not much.

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Here, wherever I have an s, I have an s minus a here.

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So if this is a function of s, what's this going to be?

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It's going to be that same function.

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Whatever the Laplace Transform of f was, it's going to be

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that same function, but instead of s, it's going to be

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a function of s minus a.

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And once again, how did I get that?

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Well I said the Laplace Transform of f is a function

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of s, and it's equal to this.

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Well if I just replace an s with an s minus a, I get this,

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which is a function of s minus a.

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Which was the Laplace Transform of e to the

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at times f of t.

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Maybe that's a little confusing.

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Let me show you an example.

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Let's just take the Laplace Transform of cosine of 2t.

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We've shown is equal to-- well I'll write the notation-- it's

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equal to some function of s.

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And that function of s is s over s squared plus 4.

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We've shown that already.

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And so the Laplace Transform of e to the, I don't know, 3t

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times cosine of 2t, is going to be equal to the same

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function, but instead of s, it's going to be a

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function of s minus a.

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So s minus 3, which is equal to s minus 3 over s minus 3

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squared plus 4.

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Notice, when you just multiply something by this, either the

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3t and then or either the at, you take the Laplace Transform

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of it, you just-- it's the same thing as the Laplace

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Transform of this function, but everywhere where you had

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an s, you replace it with an s minus this a.

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Anyway, I hope I didn't confuse you too much

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with that last part.

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I think my power adaptor actually just went on.

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I hope the video keeps recording.

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I'll see you in the next one.