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In the last video, we showed that the Laplace transform of

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f prime of t is equal to s times the Laplace transform of

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our function f minus f of 0.

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Now, what we're going to do here is actually use this

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property that we showed is true.

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And use it to fill in some more of the entries in our

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Laplace transform table, that you'll probably have to

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memorize, sooner or later, if you use Laplace

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transforms a lot.

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But we already learned that the Laplace transform of sine

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of a t is equal-- and we did a very hairy integration by

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parts problems to show that that is equal to a over s

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squared plus a squared.

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So let's use these two things we know to figure out what the

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Laplace transform of cosine of a t is.

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So the Laplace transform of cosine of a

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t is equal to what?

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Well, if we assume that the Laplace transform of cosine of

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a t is the derivative of some function, what is it the

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derivative of?

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Right?

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Let me do it on the side.

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If f prime of t is equal to cosine of a t, what is a

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potential f of t?

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Well, it's the antiderivative.

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And we can just forget about the constant, because we just

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have to know nf of t for which this is true.

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So what's the antiderivative cosine of it?

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It's 1/a sine of a t.

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Right?

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So if this is f prime of t, then that is equal to s times

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the Laplace transform of its antiderivative, or 1/a sine of

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a t minus the antiderivative evaluated at 0.

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Minus 1/a sine of-- well, a times 0 is 0.

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Well, sine of 0 is 0.

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So this whole term goes away.

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This is a constant, right?

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This 1/a.

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And we showed that the Laplace

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transform is a linear operator.

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So we can take it out.

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So this is equal to s/a times the Laplace transform

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of sine of a t.

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And that is equal to s/a times a over f

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squared plus a squared.

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And the a's cancel out.

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And that was much simpler than the integration by parts we

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had to do to figure this out.

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So then we get that the Laplace transform of cosine of

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a t is equal to s over s squared plus a squared.

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And in three minutes, we filled in another table in our

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Laplace transform table.

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And now we have the two most important trig functions.

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Let's keep going.

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We haven't really done much with polynomials.

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We know a couple of things.

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We did this already.

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We know that the Laplace transform of

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1 is equal to 1/s.

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So let's see if we could use this and the fact that the

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Laplace transform of f prime is equal to s times the

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Laplace transform of f minus f of 0.

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Or another way.

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Let's rearrange this.

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If we know f, how can we figure out some Laplace

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transforms in terms of f prime and f of 0?

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So let's just rearrange this equation.

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So we get the Laplace transform of f prime.

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I could write of t, but that gets monotonous.

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Plus f of 0 is equal to s times the Laplace

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transform of f.

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Divide both sides by s.

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Let me put the Laplace transform of-- and I'm also

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going to the sides.

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So I guess the Laplace transform-- my

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l's are getting funky.

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The Laplace transform of f is equal to 1/s.

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I'm just dividing both sides by s, so 1/s times this.

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Times the Laplace transform of my derivative plus my function

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evaluated at 0.

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And let's see if we can use this and this to figure out

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some more useful Laplace transforms.

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Well what is the Laplace transform of f of

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t is equal to t?

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Well, just use this property.

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This is going to be equal to 1/s times the Laplace

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transform of the derivative.

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Well, what's the derivative of t?

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The derivative of t is 1.

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So it's the Laplace transform of 1 minus f of 0.

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When t equals 0, this becomes 0.

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Minus 0.

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So the Laplace transform of t is equal to 1/s times the

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Laplace transform of 1.

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Well that's just 1/s.

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So it's 1 over s squared minus 0.

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Interesting.

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The Laplace transform of 1 is 1/s, Laplace transform of t is

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1/s squared.

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Let's figure out what the Laplace transform

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of t squared is.

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And I'll do this one in green.

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And maybe we'll see a pattern emerge.

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The Laplace transform of t squared.

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Well, it equals 1/s times the Laplace transform of it's

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derivative.

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So what's it's derivative?

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Times the Laplace transform of 2t plus this evaluated at 0.

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Well, that's just 0.

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So this is equal to-- well we can just take

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this constant out.

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This is equal to 2/s times the Laplace transform of t.

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Well, what does that equal?

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We just solved it.

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1/s squared.

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So it's 2/s times 1/s squared.

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So it's equal to 2/s to the third.

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Fascinating.

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Well, let me just do t the third.

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And I think then you'll see the pattern.

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The pattern will emerge.

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The Laplace transform.

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And this is actually kind of fun.

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I recommend you do it.

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It's somehow satisfying.

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It's much more satisfying than integration by parts.

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So the Laplace transform of t to the third is 1/s times the

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Laplace transform of it's

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derivative, which is 3t squared.

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Which is-- take the constant out because

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it's a linear operator.

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3/s times the Laplace transform of t squared.

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So it equals what?

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What's the Laplace transform of t squared?

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It's 2/s to the third.

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So this equals 3 times 2 over what? s to the fourth.

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And you can put a t/n here and use an inductive argument to

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figure out a general formula.

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And that general formula is-- I think you

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see the pattern here.

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Whatever my exponent is, the Laplace transform has an s in

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the denominator with one larger exponent.

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And then the numerator is the factorial of my exponent.

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So in general, and this is one more entry in our Laplace

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transform table.

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The Laplace transform of t to the nth power is equal to n

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factorial over s to the n plus 1.

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That's a parenthesis.

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I guess I didn't have to write those parenthesis.

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That just confuses it.

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But anyway, when you see this in a Laplace transform table,

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it seems intimidating.

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Oh boy, I have n's and I have n factorials and all of that.

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But it's just saying with this pattern we showed, t to the

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third increase it by 1, so s the fourth, put another

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denominator and take three factorial on the numerator,

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which is 6, right?

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And that's all it is.

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So using the derivative property of Laplace transform,

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we figured out the Laplace transform of cosine of a t and

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the Laplace transform of really any polynomial, right?

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Because it's a linear operator.

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So now we know t to the nth power, t to

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the whatever power.

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And we can multiply it by a constant.

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So we know the basic trig functions.

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We know the exponential function.

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And we know how to take the Laplace transform of

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polynomials.

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See you in the next video.

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