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A lot of what we do with Laplace transforms, taking

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them and taking their inverse, it's a

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lot of pattern matching.

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And it shouldn't just be a mechanical thing, and that's

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why I've gone through the exercise of showing

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you why they work.

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But in order to just kind of make sure we don't get

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confused, I think it might be useful to review a little bit

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of everything that we've learned so far.

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So in the last video, we saw that the Laplace transform--

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well, let me just write something.

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The Laplace transform of f of t, let me just get some

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notation down, and we can write that as big capital F of

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s, and I've told you that before.

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And so given that, in the last video I showed you that if we

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have to deal with the unit step function, so if I said,

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look, the Laplace transform of the unit step function, it

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becomes 1 at some value c times some shifted function f

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of t minus c, in the last video, we saw that this is

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just equal to e to the minus cs times the Laplace transform

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of just this function right there, so times the F of s.

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And it's really important not to get this confused with

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another Laplace transform property or rule, or whatever

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you want to call it, that we figured out.

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I think it was one of the videos that I made last year,

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but if you're just following these in order I think it's

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three or four videos ago.

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And that one told us that the Laplace transform of e to the

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at, times f of t, that this is equal to-- and I want to make

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this distinction very clear.

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Here we shifted the f of t and we got just kind of a

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regular F of s.

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In this situation, when we multiply it times a e to the

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positive at, we end up shifting the actual transform.

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So this becomes F of s minus a.

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And these two rules, or properties, or whatever you

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want to call them, they're very easy to

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confuse with each other.

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So we're going to do a couple of examples that we're going

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to have to figure out which one of these two apply.

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Let's write all the other stuff that we learned as well.

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The very first thing we learned was that the Laplace

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transform of 1 was equal to 1/s.

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We know that that's a pretty straightforward one, easy to

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prove to yourself.

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And more generally, we learned that the Laplace transform of

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t to the n, where n is a positive integer, it equaled n

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factorial over s to the n plus 1.

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And then we had our trig functions

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that we've gone over.

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Let me do this in a different color.

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I'll do it right here.

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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 the Laplace transform of the cosine of at is equal to s

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over s squared plus a squared.

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And you'll be amazed by how far we can go with just what

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I've written here.

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In future videos, we're going to broaden our toolkit even

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further, but just these right here, you can already do a

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whole set of Laplace transforms and inverse Laplace

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transforms. So let's try to do a few.

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So let's say I were to give you the Laplace transform.

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And you know, this is just the hard part.

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I think you know how to solve a differential equation, if

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you know how to take the Laplace transforms

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and go back and forth.

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The hard part is just recognizing or inverting your

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Laplace transforms. So let's say we had the Laplace

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transform of some function F of s.

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Let's say it's 3 factorial over s minus 2 to the fourth.

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Now, your pattern matching, or your pattern recognition part

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of your brain, should immediately say, look, I have

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a Laplace transform of something that has a factorial

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in it, and it's over an exponent.

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This must be something related to this

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thing right here, right?

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If I just had the Laplace transform-- let me write that

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down-- the Laplace transform of-- you see a 3 factorial and

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a fourth power, so it looks like n is equal to 3.

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So if you write the Laplace transform of t to the 3, this

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rule that we showed right here, this means that it would

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be equal to 3 factorial over s to the fourth.

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Now, this thing isn't exactly this thing.

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They're not quite the same thing.

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You know, I'm doing this to instruct you, but I find

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these, when I'm actually doing them on an exam-- I remember

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when I did them when I first learned this, I would actually

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go through this step because you definitely don't want to

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make a careless mistake and you definitely want to make

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sure you have a good handle on what you're doing.

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So you're like, OK, it's something related to this, but

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what's the difference between this expression right here and

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the expression that we're trying to take the inverse

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Laplace transform of, and this one here?

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Well, we've shifted our s.

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If we call this expression right here F of s, then what's

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this expression?

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This expression right here is F of s minus 2.

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So what are we dealing with here?

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So you see here, you have a shifted F of s.

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So in this case, a would be equal to 2.

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So this is the Laplace transform of e to the at times

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our f of t.

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So let me write this down.

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This is the Laplace transform of e to the-- and what's a?

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a is what we shifted by.

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It's what we shifted by minus a, so you have a positive a,

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so e to the 2t times the actual function.

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If this was just an F of s, what would f of t be?

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Well, we figured out, it's t the 3, t to the third power.

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So the Laplace transform of this is equal to that.

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Or we could write that the inverse Laplace transform of 3

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factorial over s minus 2 to the fourth is equal to e to

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the 2t times t to the third.

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Now, if that seemed confusing to you, you

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can kind of go forward.

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Let's go the other direction, and maybe this will make it a

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little bit clearer for you.

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So let's go from this direction.

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If I have to take the Laplace transform of this thing, I'd

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say, OK, well, the Laplace transform of t to

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the third is easy.

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I think the tool isn't working right there properly.

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Let me scroll up a little bit.

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So I could write it right here.

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So if I wanted to figure out the Laplace transform of e to

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the 2t times t to the third, I'll say, well, you know, this

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e to the 2t, I remember that it shifts something.

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So if I know that the Laplace transform of t the third, this

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is an easy one.

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It's equal to 3 factorial over s to the fourth.

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That's 3 plus 1.

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Then the Laplace transform of e to the 2t times t the third

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is going to be this shifted.

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This is equal to F of s.

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Then this is going to be f of s minus 2.

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So what's F of s minus 2?

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It's going to be equal to 3 factorial over s minus 2 to

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the fourth.

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I think you're already getting an appreciation that the

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hardest thing about these Laplace transform problems are

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really kind of all of these shifts and kind of recognizing

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the patterns and recognizing what's your a, and what's your

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c, and being very careful about it so you don't make a

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careless mistake.

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And I think doing a lot of examples probably helps a lot,

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so let's do a couple of more to kind of make sure things

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really get hammered home in your brain.

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So let's try this one right here.

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This looks a little bit more complicated.

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They give us that the Laplace transform of some function is

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equal to 2 times s minus 1 times e to the minus 2s, all

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of that over s squared minus 2s plus 2.

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Now this looks very daunting.

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How do you do this?

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I have an e here.

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I have something shifted here.

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I have this polynomial in the denominator here.

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What can I do with this?

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So the first thing, when I look at these polynomials in

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the denominator, I say can I factor it somehow?

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Can I factor it fairly simply?

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And actually, in the exams that you'll find in

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differential equation class, they'll never give you

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something that's factorable into these weird numbers.

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It tends to be integers.

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So you see, OK, what two numbers?

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They have to be positive.

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When you give their product, you get 2.

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And then when you add them, you get negative 2, or they

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could both be negative.

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But there's no two easy numbers, not 1 and 2.

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None of those work.

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So if you can't factor this outright, the next idea is

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maybe we could complete the square and maybe this will

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match one of the cosine or the sine formulas.

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So how can we complete the square in this denominator?

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Well, this can be rewritten as s squared minus 2s.

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And I'm going to put a plus 2 out here.

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And you can watch, I have a bunch of videos on the

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completing of the square, if all of this

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looks foreign to you.

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And to complete the square, we just want to turn this into a

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perfect square.

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So to turn this into a perfect square-- so something when I

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add it to itself twice becomes minus 2, and so that when I

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square it, when I add it to itself twice, it becomes minus

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2, it's minus 1.

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And when I square it, it'll become plus 1.

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I can't just add plus 1 arbitrarily to some

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expression, I have to make it neutral.

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So let me subtract 1.

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I haven't changed this.

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I added 1 and I subtracted 1.

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A little bit of a primer on completing the square.

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But by doing this, I now can call this expression right

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here, I can now say that this thing is s minus 1 squared.

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And then this stuff out here, this out here is 2 minus 1.

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This is just plus 1.

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So I can rewrite my entire expression now as 2 times s

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minus 1 times e to the minus 2s-- make sure I'm not

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clipping off at the top-- e to the minus 2s, all of that over

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s minus 1 squared plus 1.

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So a couple of interesting things seem to

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be happening here.

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Let's just do a couple of test Laplace transforms. So if a

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Laplace transform of cosine of t, we know that this is equal

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to s over s squared plus 1, which this kind of looks like

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if this was an s and this was an s squared plus 1.

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If this was F of s, then what is this?

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Well, let's ignore this guy right here for a little bit.

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So what is it?

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We know, actually, from the last video.

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We saw, well, what if we took the Laplace transform of e to

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the-- I'll call it 1t.

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But let's say e to the-- yeah I'll just write it e to the 1t

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times cosine of t?

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Well, then this will just shift this Laplace

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transform by 1.

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It will shift it by 1 to the right.

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Wherever you see an s, you would put an s minus a 1.

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So this will be equal to s minus 1 over s minus 1

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squared plus 1.

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We're getting close.

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We now figured out this part right here.

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Now, in the previous video, I think it was two videos ago,

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or maybe it was the last video, I forget.

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Memory fails me.

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I showed you that if you have the Laplace transform of the

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unit step function of t times some f of t shifted by some

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value of c, then that this is equal to e to the minus cs

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times F of s.

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OK, And this can get very confusing.

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This can get very confusing, so I want to be

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very careful here.

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Let's ignore all of this.

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I called this F of s before, but now I'm going to backtrack

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a little bit.

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And let's just ignore this, because I'm going to redefine

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our F of s.

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So let's just ignore that for a second.

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Let's define our new f of t to be this.

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Let's say that that is f of t.

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Let's say f of t is equal to e to the t cosine of t.

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Then if you take the Laplace transform of that, that means

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that F of s is equal to s minus 1 over s minus 1

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squared plus 1.

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Nothing fancy there.

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I just defined our f of t as this, and then

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our F of s is that.

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00:13:44,399 --> 00:13:48,389
Now, we have a situation here.

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Let's ignore the 2 here.

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The 2 is just kind of a scaling factor.

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This expression right here, we can rewrite as that expression

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00:13:57,179 --> 00:13:59,609
is equal to-- this is our F of s.

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00:13:59,610 --> 00:14:05,600
This expression right here is equal to 2 times our F of s

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00:14:05,600 --> 00:14:09,149
times e to the minus 2s.

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00:14:09,149 --> 00:14:10,230
Or let me just write it.

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00:14:10,230 --> 00:14:13,440
Let me switch the order, just so we make it look right.

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00:14:13,440 --> 00:14:19,140
2 times e to the minus 2s times F of s.

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00:14:19,139 --> 00:14:22,720
Well, that looks just like this if our 2 was

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equal to our c.

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00:14:23,970 --> 00:14:27,000


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00:14:27,000 --> 00:14:28,570
So what does that tell us?

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That tells us that the inverse Laplace transform, if we take

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00:14:31,590 --> 00:14:33,660
the inverse Laplace transform-- and

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let's ignore the 2.

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Let's do the inverse Laplace transform of the whole thing.

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The inverse Laplace transform of this thing is going to be

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equal to-- we can just write the 2 there as a scaling

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00:14:47,070 --> 00:14:52,000
factor, 2 there times this thing times

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00:14:52,000 --> 00:14:55,009
the unit step function.

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What's our c?

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You can just pattern match.

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00:14:57,649 --> 00:14:58,329
You have a 2 here.

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You have a c, a minus c, a minus 2, so c is 2.

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00:15:01,779 --> 00:15:06,379
The unit step function is zero until it gets to 2 times t, or

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00:15:06,379 --> 00:15:12,179
of t, so, then it becomes 1 after t is equal to 2, times

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00:15:12,179 --> 00:15:19,109
our function shifted by 2.

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00:15:19,110 --> 00:15:21,990
So this is our inverse Laplace transform.

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00:15:21,990 --> 00:15:24,950
Now, what was our function?

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Our function was this thing right here.

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00:15:29,139 --> 00:15:32,090
So if our inverse Laplace transform of that thing that I

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00:15:32,090 --> 00:15:38,170
had written is this thing, an f of t, f of t is equal to e

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00:15:38,169 --> 00:15:41,649
to the t cosine of t.

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Then our inverse-- let me write all of this down.

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Let me write our big result.

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00:15:47,179 --> 00:15:51,469
We established that the inverse Laplace transform of

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00:15:51,470 --> 00:15:55,300
that big thing that I had written before, 2 times s

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00:15:55,299 --> 00:16:05,889
minus 1 times e to the minus 2-- sorry, e to the minus 2s

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00:16:05,889 --> 00:16:13,819
over s squared minus 2s plus 2 is equal to this thing where f

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00:16:13,820 --> 00:16:14,770
of t is this.

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00:16:14,769 --> 00:16:20,699
Or we could just rewrite this as 2 times the unit step

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00:16:20,700 --> 00:16:24,129
function starting at 2, where that's when it becomes

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00:16:24,129 --> 00:16:28,629
non-zero of t times f of t minus 2. f of t minus 2 is

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00:16:28,629 --> 00:16:32,189
this with t being replaced by t minus 2.

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00:16:32,190 --> 00:16:35,120
I'll do it in another color, just to ease the monotony.

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00:16:35,120 --> 00:16:43,919
So it would be e to the t minus 2 cosine of t minus 2.

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00:16:43,919 --> 00:16:45,909
Now, you might be thinking, Sal, you know, he must have

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00:16:45,909 --> 00:16:47,990
taken all these baby steps with this problem, because

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00:16:47,990 --> 00:16:49,789
he's trying to explain it to me.

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00:16:49,789 --> 00:16:53,370
But I'm taking baby steps with this problem so that I myself

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00:16:53,370 --> 00:16:54,320
don't get confused.

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00:16:54,320 --> 00:16:56,750
And I think it's essential that you do

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take these baby steps.

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00:16:58,110 --> 00:17:00,320
And let's just think about what baby steps we took.

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00:17:00,320 --> 00:17:01,350
And I really want to review this.

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00:17:01,350 --> 00:17:02,800
This is actually a surprisingly good problem.

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00:17:02,799 --> 00:17:05,930
I didn't realize it when I first decided to do it.

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00:17:05,930 --> 00:17:07,019
We solved this thing.

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00:17:07,019 --> 00:17:09,930
We wanted to get this denominator into some form

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00:17:09,930 --> 00:17:14,560
that is vaguely useful to us, so I completed the square

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00:17:14,559 --> 00:17:17,940
there and then we rewrote our Laplace transform,

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00:17:17,940 --> 00:17:19,769
our f of s like this.

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00:17:19,769 --> 00:17:21,338
And then we used a little pattern recognition.

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00:17:21,338 --> 00:17:23,868
We said, look, if I take the Laplace transform of cosine of

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00:17:23,868 --> 00:17:27,750
t, I'd get s over s squared plus 1.

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00:17:27,750 --> 00:17:29,500
But this isn't s over s squared plus 1.

325
00:17:29,500 --> 00:17:33,180
It's s minus 1 over s minus 1 squared plus 1.

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00:17:33,180 --> 00:17:36,450
So we said, oh, well, that means that we're multiplying

327
00:17:36,450 --> 00:17:38,299
our original time domain function.

328
00:17:38,299 --> 00:17:43,849
We're multiplying our f of t times e to the 1t.

329
00:17:43,849 --> 00:17:46,199
And that's what we got there.

330
00:17:46,200 --> 00:17:49,460
So the Laplace transform of e to the t cosine of t became s

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00:17:49,460 --> 00:17:53,180
minus 1 over s minus 1 squared plus 1.

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00:17:53,180 --> 00:17:56,970
And then we had this e to the minus 2s this entire time.

333
00:17:56,970 --> 00:18:01,470
And that's where we said, hey, if we have e to the minus 2s

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00:18:01,470 --> 00:18:04,610
in our Laplace transform, when you take the inverse Laplace

335
00:18:04,609 --> 00:18:09,759
transform, it must be the step function times the shifted

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00:18:09,759 --> 00:18:11,019
version of that function.

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00:18:11,019 --> 00:18:12,200
And that's why I was very careful.

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00:18:12,200 --> 00:18:14,319
And you had this 2 hanging out the whole time, and I could

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00:18:14,319 --> 00:18:15,029
have used that any time.

340
00:18:15,029 --> 00:18:19,710
But the simple constants just scale.

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00:18:19,710 --> 00:18:21,960
A function is equal to two times the Laplace transform of

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00:18:21,960 --> 00:18:23,410
that function and vice versa.

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00:18:23,410 --> 00:18:25,250
So the 2's are very easy to deal with, so I kind of

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00:18:25,250 --> 00:18:27,470
ignored that most of the time.

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00:18:27,470 --> 00:18:28,779
But that's why I was very careful.

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00:18:28,779 --> 00:18:34,420
I redefined f of t to be this, F of s to be this, and said,

347
00:18:34,420 --> 00:18:39,920
gee, if F of s is this, and if I'm multiplying it times e to

348
00:18:39,920 --> 00:18:45,039
the minus 2s, then what I'm essentially doing, I'm fitting

349
00:18:45,039 --> 00:18:46,730
this pattern right here.

350
00:18:46,730 --> 00:18:49,069
And so the answer to my problem is going to be the

351
00:18:49,069 --> 00:18:52,450
unit step function-- I just throw the 2 out there-- the 2

352
00:18:52,450 --> 00:18:57,960
times the unit step function times my f of t shifted by c.

353
00:18:57,960 --> 00:19:00,360
And we established this was our f of t, so we just

354
00:19:00,359 --> 00:19:02,500
shifted it by c.

355
00:19:02,500 --> 00:19:05,569
We shifted it by 2, and we got our final answer.

356
00:19:05,569 --> 00:19:09,220
So this is about as hard up to this point as you'll see an

357
00:19:09,220 --> 00:19:11,779
inverse Laplace transform problem.

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00:19:11,779 --> 00:19:14,450
So, hopefully, you found that pretty interesting.

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00:19:14,450 --> 00:19:14,615