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Welcome back.

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We were in the midst of figuring out the Laplace

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transform of sine of at when I was running out of time.

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This was the definition of the Laplace

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transform of sine of at.

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I said that also equals y.

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This is going to be useful for us, since we're going to be

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doing integration by parts twice.

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So I did integration by parts once, then I did integration

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by parts twice.

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I said, you know, don't worry about the boundaries of the

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integral right now.

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Let's just worry about the indefinite integral.

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And then after we solve for y-- let's just say y is the

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indefinite version of this-- then we can evaluate the

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boundaries.

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And we got to this point, and we made the realization, after

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doing two integration by parts and being very careful not to

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hopefully make any careless mistakes, we realized, wow,

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this is our original y.

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If I put the boundaries here, that's the same thing as the

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Laplace transform of sine of at, right?

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That's our original y.

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So now-- and I'll switch colors just avoid monotony--

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this is equal to, actually, let me just-- this is y.

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Right?

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That was our original definition.

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So let's add a squared over sine squared y to

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both sides of this.

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So this is equal to y plus-- I'm just adding this whole

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term to both sides of this equation-- plus a squared over

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s squared y is equal to-- so this term is now gone, so it's

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equal to this stuff.

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And let's see if we can simplify this.

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So let's factor out an e to the minus st. Actually, let's

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factor out a negative e to the minus st. So it's minus e to

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the minus st, times sine of-- well, let me just write 1 over

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s, sine of at, minus 1 over s squared, cosine of at.

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I really hope I haven't made any careless mistakes.

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And so this, we can add the coefficient.

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So we get 1 plus a squared, over s squared, times y.

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But that's the same thing as s squared over s squared, plus a

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squared over s squared.

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So it's s squared plus a squared, over s squared, y is

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equal to minus e to the minus st, times this whole thing,

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sine of at, minus 1 over s squared, cosine of at.

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And now, this right here, since we're doing everything

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with respect to dt, this is just a constant, right?

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So we can say a constant times the

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antiderivative is equal to this.

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This is as good a time as any to evaluate the boundaries.

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Right?

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If this had a t here, I would have to somehow get them back

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on the other side.

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Because the t's are involved in evaluating the boundaries,

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since we're doing our definite integral or improper integral.

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So let's evaluate the boundaries now.

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And we could've kept them along with us

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the whole time, right?

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And just factored out this term right here.

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But anyway.

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So let's evaluate this from 0 to infinity.

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And this should simplify things.

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So the right-hand side of this equation, when I evaluate it

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at infinity, what is e to the minus infinity?

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Well, that is 0.

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We've established that multiple times.

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And now it approaches 0 from the negative side, but it's

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still going to be 0, or it approaches 0.

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What's sine of infinity?

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Well, sine just keeps oscillating, between negative

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1 and plus 1, and so does cosine.

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Right?

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So this is bounded.

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So this thing is going to overpower these.

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And if you're curious, you can graph it.

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This kind of forms an envelope around these oscillations.

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So the limit, as this approaches infinity, is going

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to be equal to 0.

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And that makes sense, right?

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These are bounded between 0 and negative 1.

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And this approaches 0 very quickly.

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So it's 0 times something bounded between 1

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and negative 1.

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Another way to view it is the largest value this could equal

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is 1 times whatever coefficient's on it, and then

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this is going to 0.

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So it's like 0 times 1.

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Anyway, I don't want to focus too much on that.

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You can play around with that if you like.

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Minus this whole thing evaluated at 0.

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So what's e to the minus 0?

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Well, e to the minus 0 is 1.

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Right?

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That's e to the 0.

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We have a minus 1, so it becomes plus 1 times-- now,

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sine of 0 is 0.

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Minus 1 over s squared, cosine of 0.

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Let's see.

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Cosine of 0 is 1, so we have minus 1 over s squared, minus

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1 over s squared, times 1.

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So that is equal to minus 1 over s squared.

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And I think I made a mistake, because I shouldn't be having

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a negative number here.

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So let's backtrack.

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Maybe this isn't a negative number?

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Let's see, infinity, right?

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This whole thing is 0.

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When when you put 0 here, this becomes a minus 1.

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Yeah.

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So either this is a plus or this is a plus.

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Let's see where I made my mistake.

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e to the minus st-- oh, I see where my mistake is.

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Right up here.

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Where I factored out a minus e to the minus st, right?

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Fair enough.

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So that makes this 1 over s, sine of at.

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But if I factor out a minus e to the minus st, this becomes

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a plus, right?

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It was a minus here, but I'm factoring out of a minus e to

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the minus st.

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So that's a plus.

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This is a plus.

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Boy, I'm glad that was not too difficult to find.

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So then this becomes a plus.

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And then this becomes a plus.

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Thank God.

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It would have been sad if I wasted two videos and ended up

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with a careless, negative number.

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Anyway.

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So now we have s squared plus a squared, over s squared,

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times y is equal to this.

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Multiply both sides times s squared over-- s

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squared plus a squared.

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Divide both sides by this, and we get y is equal to 1 over s

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squared-- And actually, let me make sure that that is right.

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It's 1 over s squared.

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y is equal to 1 over s squared, times s squared, over

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s squared plus a squared.

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And then these cancel out.

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And let me make sure that I haven't made

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another careless mistake.

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Because I have a

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feeling I have. Yep.

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There.

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I see the careless mistake.

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And it was all in this term.

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And I hope you don't mind my careless mistakes, but I want

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you to see that I'm doing these things in real time and

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I am human, in case you haven't realized already.

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Anyway, so I made the same careless mistake.

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So I factor out an e to the minus st here, so it's plus.

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But it was a over s squared.

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So this is an a.

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That's an a.

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And so this is an a.

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And so this is an a.

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And so this is an a.

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Right?

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This was an a.

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And this is the correct answer.

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a over s squared plus a squared.

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So I hope those careless mistakes didn't

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throw you off too much.

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These things happen when you do integration by parts twice

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with a bunch of variables.

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But anyway, now we are ready to add a significant entry

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into our table of Laplace transforms. And that is that

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the Laplace transform-- I had an extra curl, there.

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That was unecessary.

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Let me do it again.

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The Laplace transform of sine of at is equal to a over s

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squared, plus a squared.

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And that's a significant entry.

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And maybe a good exercise for you, just to see how fun it is

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to do these integration by parts problems twice, is to

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figure out the Laplace transform of cosine of at.

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And I'll give you a hint.

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It's s over s squared over s squared plus a squared.

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And it's nice that there's that symmetry there.

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Anyway, I'm almost at my time limit.

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And I'm very tired working on this video.

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So I'll leave it there and I'll see you in the next one.

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