1
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Now that we know a little bit about the convolution integral

3
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and how it applies to the Laplace transform, let's

4
00:00:05,200 --> 00:00:07,950
actually try to solve an actual differential equation

5
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using what we know.

6
00:00:09,269 --> 00:00:12,439
So I have this equation here, this initial value problem,

7
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where it says that the second derivative of y plus 2 times

8
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the first derivative of y, plus 2 times y, is equal to

9
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sine of alpha t.

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And they give us some initial conditions.

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They tell us that y of 0 is equal to 0, and that y prime

12
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of 0 is equal to 0.

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And that's nice and convenient that those initial conditions

14
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tend to make the problem pretty clean.

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But let's get to the problem.

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So the first thing we do is we take the Laplace transform of

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both sides of this equation.

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The Laplace transform of the second derivative

19
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of y is just s squared.

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This should be a bit of second nature to you by now.

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It's s squared times the Laplace transform of Y, which

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I'll just write as capital Y of s, minus s-- so we start

23
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with the same degree as the number of derivatives we're

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taking, and then we decrement that every time-- minus s

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00:01:12,040 --> 00:01:16,130
times y of 0-- you kind of think of this as the integral,

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and you take the derivative 1, so this isn't exactly the

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derivative of that-- minus, you decrement that 1, you just

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have a 1 there, y prime of o.

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And that's the Laplace transform of the second

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derivative.

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Now, we have to do the Laplace transform of 2 times the first

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derivative.

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That's just going to be equal to plus 2, times sY of s-- s

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times the Laplace transform of Y; that's that

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there-- minus y of 0.

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And we just have one left.

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The Laplace transform of 2Y.

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That's just equal to plus 2 times the Laplace

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transform of Y.

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And then that's going to be equal to the Laplace transform

41
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of sine of alpha t.

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We've done that multiple times so far.

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That's just alpha over s squared plus alpha squared.

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Now, the next thing we want to do is we want to separate out

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the Laplace transform of Y terms, or the Y of s terms.

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Actually, even better, let's get rid of these initial

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conditions.

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y of 0, and y prime of 0 is 0, so this term is 0.

49
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That term is 0, and that term is 0.

50
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So our whole expression-- I can get rid of the colors

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now-- it just becomes-- let me pick a nice color here--

52
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becomes s squared times Y of s, plus 2s, Y of s-- that's

53
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that term right there-- plus 2Y of s, is equal to the

54
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right-hand side, is equal to alpha over s squared plus

55
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alpha squared.

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Now let's factor out the Y of s, or the Laplace

57
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transform of Y.

58
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And so we get s squared plus 2s, plus 2, all of that times

59
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Y of s, is equal to this right-hand side, is equal to

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alpha over s squared, plus alpha squared.

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Now we can divide both sides of this equation by this thing

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right here, by that right there.

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And we get Y of s, the Laplace transform of Y is equal to

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this thing, alpha over s squared, plus alpha squared,

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times-- or, you know, I could just say times-- 1 over s

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squared, plus 2s, plus 2.

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I could just say divided by this, but it works out the

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same either way.

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Now, what can we do here?

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Remember, I was doing this in the context of convolution, so

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I want to look for a Laplace transform that looks like the

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product of two Laplace transforms. I know what the

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inverse Laplace transform of this is.

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In fact, I just took it.

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It's sine of alpha t.

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So if I can figure out the inverse Laplace transform of

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this, I could at least express our function y of t at least

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as a convolution integral, even if I don't necessarily

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solve the integral.

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From there, it's just calculus, or if it's an

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unsolvable integral, we could just use a computer or

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something, although you could actually use a computer to

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solve this so, you might skip some steps even

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going through this.

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But anyway, let's just try to get this in terms of a

86
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convolution integral.

87
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So what can I do with this?

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This is, let's see, this isn't a perfect square.

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So if this isn't a perfect square, the next best thing is

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to try to complete the square here.

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So let's try to write this as a, so let's see, if I write

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this as s squared plus 2s, plus something, plus 2.

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I just rewrote it like this.

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And if I wrote this as s squared plus 2s, plus 1, that

95
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becomes s plus 1 squared.

96
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But if I add a 1, I have to also subtract a 1.

97
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I can't just add 1's arbitrarily to things.

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So if I add 1 I have to subtract a 1 to cancel out

99
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with that 1.

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So I really haven't changed this at all, I just

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rewrote it like this.

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But this now, I can rewrite this term right

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here as s plus 1 squared.

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And then this becomes plus 1.

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That's this term right here.

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This is the plus 1.

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So I could rewrite my whole Y of s is now equal to alpha

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over s squared, plus alpha squared, times 1 over this

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thing, s plus 1 squared, plus 1.

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Now, I already said, I know what the inverse Laplace

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transform of this thing is.

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Now I just have to figure out what the inverse Laplace

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transform of this thing is.

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Of this-- let me pick a nice color-- of this blue thing in

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the blue box, and then I can express it as

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a convolution integral.

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And how do I do that?

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I could just do it right now.

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I could just immediately say that y of t-- let me write

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this down-- y of t, so the inverse is equal to the

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inverse Laplace transform of, obviously of Y of s.

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Let me write that down, Y of s.

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Which is equal to the inverse Laplace transform

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of these two things.

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The inverse Laplace transform of alpha over s squared, plus

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alpha squared, times 1 over s plus 1 squared, plus 1.

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And now the convolution theorem tells us that this is

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going to be equal to the inverse Laplace transform of

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this first term in the product.

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So the inverse Laplace transform of that first term,

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alpha over s squared, plus alpha squared, convoluted

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with-- I'll do a little convolution sign there.

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I was about to say convulsion.

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They're not too different.

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Convoluted with the inverse Laplace transform of this

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term, the inverse Laplace transform of 1 over s plus 1

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squared, plus 1.

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If I have the product of two Laplace transforms, and I can

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take each of them independently and I can invert

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them, the inverse Laplace transform of their product is

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going to be the convolution of the inverse Laplace transforms

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of each of them, each of the terms.

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And what I just said confused me a bit, so I don't want to

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confuse you.

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But I think you get the idea.

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I have these two things.

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I recognize these independently.

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I can independently take the inverse of each of these

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things, so the inverse Laplace transform of their products is

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00:08:02,839 --> 00:08:07,929
going to be the convolution of each of their inverse

152
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transforms. Now what's this over here?

153
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Well I had this in the beginning of the problem?

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The inverse Laplace transform of this, right here, is sine

155
00:08:16,209 --> 00:08:17,459
of alpha t.

156
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157
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And then we're going to convolute that with the

158
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inverse Laplace transform of this right here.

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Let's do a little bit of work on the side, just to make sure

160
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we get this right.

161
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So the Laplace transform of sine of t is equal to 1 over s

162
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squared, plus 1.

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That looks like this, but I was shifted by minus 1.

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You might remember that the Laplace transform of e to the

165
00:08:52,779 --> 00:08:58,089
at sine of t, when you multiply e to the at times

166
00:08:58,090 --> 00:09:00,350
anything, you're shifting its Laplace transform.

167
00:09:00,350 --> 00:09:04,820
So that will be equal to 1 over s minus a

168
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squared, plus 1.

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00:09:08,899 --> 00:09:10,389
We essentially shifted it by a.

170
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So now we have something that looks very similar to this.

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If we just set our a to be equal to negative 1, here our

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a is equal to negative 1, then it fits this pattern.

173
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This is s minus negative 1.

174
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So the inverse Laplace transform of this thing right

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here is just e to the a, which is minus 1, so minus 1t,

176
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times sine of t.

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So this is the solution to our differential equation, even

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though it's not in a pleasant form to look at.

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And we can, if we want, express it as an integral.

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I'm not going to actually solve the integral in this

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00:09:50,409 --> 00:09:52,569
problem, because it gets hairy, and it's not even clear

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that-- well, I won't even attempt to do it.

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But I just want to get into a form, and from there it's just

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00:09:56,100 --> 00:09:58,019
integral calculus.

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00:09:58,019 --> 00:10:01,500
or maybe a computer.

186
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What's the convolution of these two things?

187
00:10:03,110 --> 00:10:10,399
It's the integral from 0 to t, of sine of the first function

188
00:10:10,399 --> 00:10:11,809
of t minus tau.

189
00:10:11,809 --> 00:10:14,599


190
00:10:14,600 --> 00:10:16,800
Well, I could actually switch, and I haven't shown you this,

191
00:10:16,799 --> 00:10:20,169
but we can switch the order either way, but actually let

192
00:10:20,169 --> 00:10:21,419
me just do it this way.

193
00:10:21,419 --> 00:10:24,719


194
00:10:24,720 --> 00:10:27,389
I could write this as sine of [? out ?]

195
00:10:27,389 --> 00:10:32,949
t minus tau, times alpha-- I'm taking the sine of all of

196
00:10:32,950 --> 00:10:42,870
those things-- times e to the minus tau, sine of tau, dtau.

197
00:10:42,870 --> 00:10:45,330
That's one way, that if I wanted to express the solution

198
00:10:45,330 --> 00:10:47,310
of this differential equation's integral, I could

199
00:10:47,309 --> 00:10:48,309
write it like that.

200
00:10:48,309 --> 00:10:50,659
And it actually should be kind of obvious to you that this

201
00:10:50,659 --> 00:10:51,889
could go either way.

202
00:10:51,889 --> 00:10:57,189
Because when it was a product up here, obviously order does

203
00:10:57,190 --> 00:10:57,730
not matter.

204
00:10:57,730 --> 00:11:00,470
I could write this term first, or I could write that term

205
00:11:00,470 --> 00:11:05,450
first. So regardless of which term is written first, the

206
00:11:05,450 --> 00:11:07,690
same principle would apply.

207
00:11:07,690 --> 00:11:09,910
And I'll formally prove it in a future video.

208
00:11:09,909 --> 00:11:12,289
So we could have also done it the other way.

209
00:11:12,289 --> 00:11:16,259
We could have written this expression as e to the minus

210
00:11:16,259 --> 00:11:22,700
t, sine of t, convoluted with sine of alpha t.

211
00:11:22,700 --> 00:11:27,509
And that would be equal to the integral from 0 to t, of e to

212
00:11:27,509 --> 00:11:38,559
the minus t minus tau, sine of t, minus tau, times sine of

213
00:11:38,559 --> 00:11:40,799
alpha tau, dtau.

214
00:11:40,799 --> 00:11:42,769
So these are equivalent.

215
00:11:42,769 --> 00:11:46,439
Either of these would be an acceptable answer.

216
00:11:46,440 --> 00:11:48,480
And normally on a test like this, the teacher won't expect

217
00:11:48,480 --> 00:11:50,240
you to actually evaluate these integrals.

218
00:11:50,240 --> 00:11:53,370
The teacher will just say, get it into an integral just to

219
00:11:53,370 --> 00:11:56,149
kind of see whether you know how to convolute things and

220
00:11:56,149 --> 00:11:59,259
get your solution to the differential equation at least

221
00:11:59,259 --> 00:12:02,309
into this form, because from here it really is just, I

222
00:12:02,309 --> 00:12:03,859
won't say basic calculus, but it's

223
00:12:03,860 --> 00:12:05,134
non-differential equations.

224
00:12:05,134 --> 00:12:08,559
So hopefully, this second example with the convolution

225
00:12:08,559 --> 00:12:10,789
to solve an inverse transform clarified

226
00:12:10,789 --> 00:12:12,939
things up a little bit.

227
00:12:12,940 --> 00:12:13,209