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I'll now introduce you to the concept

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of the Laplace Transform.

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And this is truly one of the most useful concepts that

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you'll learn, not just in differential equations, but

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really in mathematics.

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And especially if you're going to go into engineering, you'll

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find that the Laplace Transform, besides helping you

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solve differential equations, also helps you transform

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functions or waveforms from the time domain to this

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frequency domain, and study and understand a

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whole set of phenomena.

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But I won't get into all of that yet.

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Now I'll just teach you what it is.

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Laplace Transform.

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I'll teach you what it is, make you comfortable with the

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mathematics of it and then in a couple of videos from now,

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I'll actually show you how it is useful to use it to solve

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differential equations.

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We'll actually solve some of the differential equations we

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did before, using the previous methods.

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But we'll keep doing it, and we'll solve more and more

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difficult problems.

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So what is the Laplace Transform?

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Well, the Laplace Transform, the notation is the L like

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Laverne from Laverne and Shirley.

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That might be before many of your times, but

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I grew up on that.

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Actually, I think it was even reruns when I was a kid.

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So Laplace Transform of some function.

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And here, the convention, instead of saying f of x,

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people say f of t.

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And the reason is because in a lot of the differential

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equations or a lot of engineering you actually are

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converting from a function of time to

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a function of frequency.

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And don't worry about that right now.

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If it confuses you.

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But the Laplace Transform of a function of t.

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It transforms that function into some other function of s.

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and And does it do that?

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Well actually, let me just do some mathematical notation

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that probably won't mean much to you.

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So what does it transform?

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Well, the way I think of it is it's kind of

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a function of functions.

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A function will take you from one set of-- well, in what

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we've been dealing with-- one set of numbers to another set

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of numbers.

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A transform will take you from one set of functions to

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another set of functions.

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So let me just define this.

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The Laplace Transform for our purposes is defined as the

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improper integral.

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I know I haven't actually done improper integrals just yet,

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but I'll explain them in a few seconds.

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The improper integral from 0 to infinity of e to the minus

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st times f of t-- so whatever's between the Laplace

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Transform brackets-- dt.

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Now that might seem very daunting to you and very

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confusing, but I'll now do a couple of examples.

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So what is the Laplace Transform?

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Well let's say that f of t is equal to 1.

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So what is the Laplace Transform of 1?

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So if f of t is equal to 1-- it's just a constant function

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of time-- well actually, let me just substitute exactly the

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way I wrote it here.

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So that's the improper integral from 0 to infinity of

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e to the minus st times 1 here.

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I don't have to rewrite it here, but there's a times 1dt.

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And I know that infinity is probably bugging you right

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now, but we'll deal with that shortly.

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Actually, let's deal with that right now.

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This is the same thing as the limit.

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And let's say as A approaches infinity of the integral from

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0 to Ae to the minus st. dt.

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Just so you feel a little bit more comfortable with it, you

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might have guessed that this is the same thing.

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Because obviously you can't evaluate infinity, but you

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could take the limit as something approaches infinity.

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So anyway, let's take the anti-derivative and evaluate

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this improper definite integral, or

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this improper integral.

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So what's anti-derivative of e to the minus st

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with respect to dt?

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Well it's equal to minus 1/s e to the minus st, right?

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If you don't believe me, take the derivative of this.

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You'd take minus s times that.

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That would all cancel out, and you'd just be left with e to

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the minus st. Fair enough.

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Let me delete this here, this equal sign.

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Because I could actually use some of that real estate.

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We are going to take the limit as A approaches infinity.

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You don't always have to do this, but this is the first

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time we're dealing with improper intergrals.

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So I figured I might as well remind you that

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we're taking a limit.

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Now we took the anti-derivative.

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Now we have to value it at A minus the anti-derivative

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valued at 0.

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And then take the limit of whatever that ends up being as

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A approaches infinity.

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So this is equal to the limit as A approaches infinity.

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OK.

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If we substitute A in here first, we get minus 1/s.

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Remember we're, dealing with t.

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We took the integral with respect to t.

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e to the minus sA, right?

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That's what happens when I put A in here.

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Now what happens when I put t equals 0 in here?

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So when t equals 0, it becomes e to the minus s times 0.

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This whole thing becomes 1.

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And I'm just left with minus 1/s.

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Fair enough.

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And then let me scroll down a little bit.

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I wrote a little bit bigger than I wanted

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to, but that's OK.

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So this is going to be the limit as A approaches infinity

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of minus 1/s e to the minus sA minus 1/s.

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So plus 1/s.

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So what's the limit as A approaches infinity?

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Well what's this term going to do?

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As A approaches infinity, if we assume that s is greater

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than 0-- and we'll make that assumption for now.

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Actually, let me write that down explicitly.

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Let's assume that s is greater than 0.

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So if we assume that s is greater than 0, then as A

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approaches infinity, what's going to happen?

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Well this term is going to go to 0, right? e to the minus--

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a googol is a very, very small number.

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And an e to the minus googol is an even smaller number.

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So then this e to the minus infinity approaches 0, so this

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term approaches 0.

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This term isn't affected because it has no A in it, so

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we're just left with 1/s.

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So there you go.

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This is a significant to moment in your life.

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You have just been exposed to your first Laplace Transform.

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I'll show you in a few videos, there are whole tables of

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Laplace Transforms, and eventually we'll

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prove all of them.

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But for now, we'll just work through some of

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the more basic ones.

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But this can be our first entry in our

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Laplace Transform table.

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The Laplace Transform of f of t is equal to

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1 is equal to 1/s.

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Notice we went from a function of t-- although obviously this

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one wasn't really dependent on t-- to a function of s.

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I have about 3 minutes left, but I don't think that's

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enough time to do another Laplace Transform.

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So I will save that for the next video.

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See you soon.

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