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Let's try to fill in our Laplace transform table a

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little bit more.

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And a good place to start is just to write our definition

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of the Laplace transform.

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The Laplace transform of some function f of t is equal to

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the integral from 0 to infinity, of e to the minus

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st, times our function, f of t dt.

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That's our definition.

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The very first one we solved for-- we could even do it on

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the side right here-- was the Laplace transform of 1.

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You know, we could almost view that as t to the 0, and that

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was equal to the integral from 0 to infinity.

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f of t was just 1, so it's e to the minus st, dt, which is

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equal to the antiderivative of e to the minus st, which is

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minus 1 over s, e to the minus st. And then you have to

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evaluate that from 0 to infinity.

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When you take the limit as this term approaches infinity,

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this e to the minus, this becomes e to the minus

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infinity, if we assume s is greater than 0.

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So if we assume s is greater than 0, this whole

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term goes to 0.

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So you end up with a 0 minus this thing evaluated at 0.

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So when you evaluate t is equal to 0, this term right

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here becomes 1, e to the 0 becomes 1, so it's minus minus

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1/s, which is the same thing as plus 1/s.

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the?

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Laplace transform of 1, of just the constant

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function 1, is 1/s.

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We already solved that.

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Now, let's increment it a little bit.

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Let's see if we can figure out the Laplace transform of t.

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So we can view this as t to the 0.

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Now this is t to the 1.

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This is going to be equal to the integral from 0 to

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infinity, of e to the minus st times t dt.

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Now.

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I can tell you right now that I don't have the

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antiderivative of this memorized.

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I don't know what it is.

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But there's a sense that the integration by parts could be

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useful, because integration by parts kind of decomposes into

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a simpler problem.

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And I always forget integration by parts, so I'll

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rederive it here in this purple color.

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So if we have u times v, if we take the derivative with

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respect to t of that, that's equal to the derivative of the

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first times the second function plus the first

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function times the derivative of the second function.

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It's just the product rule.

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Now, if we take the integral of both sides of this

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equation, we get uv is equal to the antiderivative of u

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prime v plus the antiderivative of uv prime.

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Now, since we want to apply this to an integral, maybe

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let's make this what we want to solve for, so we can get

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the integral of uv prime is equal to-- we can just

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subtract this from that side of the equation, so it's

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equal-- I'm just swapping the sides, so I'm just solving for

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this, and to solve for this, I just subtract this from that,

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so it's equal to uv minus the integral of u prime v.

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So there you go.

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Even though I have trouble memorizing this formula, it's

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not too hard to rederive as long as you remember the

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product rule right there.

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So if we're going to do integration by parts, it's

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good to define our v prime to be something that's easy to

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take the antiderivative of, because we're going to have to

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figure out v later on, and it's good to take u to be

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something that's easy take the derivative of.

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So let's make t is equal to our u and let's make e to the

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minus st as being our v prime.

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If that's the case, then what is v?

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Well, v's just the antiderivative of that.

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In fact, we've done it before.

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It's minus 1/s, e to the minus st. That's v.

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And then if we want to figure out u prime, because we're

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going to have to figure out that later anyway, u prime's

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just the derivative of t.

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That's just equal to 1.

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So let's apply this.

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Let's see, so the Laplace transform of t is equal to uv.

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u is t, v is this right here.

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It's times minus 1/s, e to the minus st. e to the minus st,

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that's the uv term right there.

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And this is a definite integral, right?

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So we're going to evaluate this term right

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here from 0 to infinity.

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And then it's minus the integral from 0 to infinity of

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u prime, which is just 1 times v.

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v, we just figured out here, is minus-- let me write it in

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v's color-- times minus 1/s-- this is my v right here--

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minus 1/s, e to the minus st, dt.

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All of that is dt.

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So let's see if we can simplify this.

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So this is equal to minus t/s, e to the minus st, evaluated

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from 0 to infinity.

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And let's see, we could take-- well, this is just 1.

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1 times anything is 1.

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We can just not write that.

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And then bring the minus 1/s out.

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So if we bring the minus 1/s out, this becomes plus 1/s

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times the integral from 0 to infinity of e to

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the minus st, dt.

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And this should look familiar to you.

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This is exactly what we solved for right here.

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It was the Laplace transform of 1.

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So let's keep that in mind.

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So this right here is the Laplace transform of 1.

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And I want to write it that way, because we're going to

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see a pattern of this in the next video.

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I'm going to write that as a Laplace transform of 1.

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But what is this equal to?

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So we're going to evaluate this as it adds to infinity

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and then subtract from that evaluated at 0.

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You can kind of view it as a substitution, so this is equal

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to-- well, let me write it this way.

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It's the limit as A approaches infinity, of minus A/s, e to

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the minus sA.

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So that's this evaluated at infinity.

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And then from that, we're going to subtract this

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evaluated at 0.

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So minus all of this, but we already have a minus sign

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here, so we could write a plus.

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Plus 0/s times e to the minus s times 0.

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And then, of course, we have this term right here.

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So let me write that term.

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I'll do it in yellow or let me do it in blue.

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Plus 1/s-- that's this right there-- times the Laplace

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transform of 1.

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And what do we get?

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So what's the limit of this as A approaches infinity?

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You might say, wow, you know, as A approaches infinity right

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here, this becomes a really big number.

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There's a minus sign in there, so it would be a really big

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negative number.

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But this is an exponent.

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A is an exponent right here.

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So e to the minus infinity is going to go to zero much

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faster than this is going to go to infinity.

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This term right here is a much stronger function, I guess is

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the way you could see it.

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And you could try it out on your calculator, if you don't

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believe me.

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This term is going to overpower this term and so

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this whole thing is going to go to zero.

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Likewise, e to the minus-- e to the 0, this is 1, but

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you're multiplying it times a zero, so this is also going to

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go to zero, which is convenient because all of this

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stuff just disappears.

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And we're left with the Laplace transform of t is

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equal to 1/s times the Laplace transform of 1.

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And we know what the Laplace transform of 1 is.

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The Laplace transform of 1-- we just did it at beginning of

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the video-- was equal to 1/s, if we assume that s is

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greater than zero.

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In fact, we have to assume that s was greater than zero

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here in order to assume that this goes to zero.

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Only if s is greater than zero, when you get a minus

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infinity here does this approach zero.

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So fair enough.

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

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which is equal to 1/s squared, where s is greater than zero.

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So we have one more entry in our table, and

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then we can use this.

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What we're going to do in the next video is build up to the

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Laplace transform of t to any arbitrary exponent.

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And we'll do this in the next video.