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Let's do another problem with repeated roots.

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So let's say our differential equation is the second

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derivative of y minus the first derivative plus 0.25--

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that's what's written here-- 0.25y is equal to 0.

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And they've actually given us some initial conditions.

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They said that y of 0 is equal to 2, and y prime of 0 is

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equal to 1/3.

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So like we've done in every one of these constant

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coefficient linear second order homogeneous differential

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equations, let's get the characteristic equation.

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So that's r squared minus r plus 0.25-- or we can even

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call it plus 1/4.

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So let's see, when I just inspected this, it always

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confuses me when I have fractions.

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So it becomes very hard to factor.

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So let's just do the quadratic formula.

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So the roots of this are going to be r is

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equal to negative b.

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Well, b is negative 1.

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So negative b is going to be 1.

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Plus or minus the square root of b squared.

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b is negative 1.

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So that squared is 1.

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Minus 4 times a, which is 1, times c.

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Well, 4 times 1 times 0.25, that's 1.

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Ah-ha.

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So notice that when you have a repeated root, this under the

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square root becomes 0.

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And that makes sense, because it's this plus or minus in the

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quadratic formula that gives you two roots, whether they be

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real or complex.

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But if the square root is 0, you're adding plus or minus 0

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and you're only left with one root.

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Anyway, we're not done yet.

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What's the denominator of a quadratic equation?

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2a.

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So a is 1, so over 2.

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So our one repeated root is 1 plus or minus 0 over 2, or it

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equals 1/2.

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And like we learned in the last video, you might just

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say, oh well, maybe the solution is just y is equal to

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ce to the 1/2 x.

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But like we pointed out last time, you have two initial

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conditions.

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And this solution is not general enough for two initial

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conditions.

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And then last time, we said, OK, if this isn't general

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enough, maybe some solution that was some function of x

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times e to the 1/2 x, maybe that would be our solution.

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And we said, it turns out it is.

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And so that more general solution that we found, that

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we figured out that v of x is actually equal to some

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constant plus x times some other constant.

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So our more general solution is y is equal to c1 times e to

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the 1/2 x soon. plus c2 times xe to the 1/2 x.

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I forgot the x here.

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Let me draw a line here so you don't get confused.

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Anyway, that's the reasoning.

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That's how we came up with this thing.

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And it is good to know.

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Because later on when you want to know more theory of

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differential equations-- and that's really the whole point

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about learning this if your whole goal isn't just to pass

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an exam-- it's good to know.

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But when you're actually solving these you could just

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kind of know the template.

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If I have a repeated root, well I just put that repeated

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root twice, and one of them gets an x in front of it, and

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they have two constants.

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Anyway, this is our general solution and now we can use

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our initial conditions to solve for c1 and c2.

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So let's just figure out the derivative of this first. So

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it becomes easy to substitute in for c2.

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So y prime is equal to 1/2 c1 e to the 1/2 x, plus-- now

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this becomes a little bit more complicated, we're going to

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have to use the product rule here-- so plus c2 times--

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derivative of x is 1-- times e to the 1/2 x, that's the

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product rule.

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Plus the derivative of e to the 1/2 x times x.

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So that's 1/2 xe to the 1/2 x.

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Or we can write-- I don't want to lose this stuff up here--

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we can write that it equals-- let's see, I have 1/2-- so I

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have c2 times e to the 1/2 x and I have 1/2 times c1

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e to the 1/2 x.

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So I could say, it's equal to e to the 1/2 x

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times c1 over 2.

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That's that.

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Plus c2.

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That takes care of these two terms. Plus c2 over 2

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xe to the 1/2 x.

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And now let's use our initial conditions.

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And let me actually clear up some space, because I think

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it's nice to have our initial conditions up here where we

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can see them.

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So let me delete all this stuff here.

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That, hopefully, makes sense to you by now.

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You know the characteristic equation.

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We figured out the general solu-- I don't want to erase

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our initial conditions-- we figured out the general

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solution was this.

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I'll keep our general solution there.

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And so, now we just substitute our initial conditions into

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our general solution and the derivative of the general

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solution, and hopefully we can get meaningful answers.

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So substituting into our general solution, y of 0 is

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equal to 2.

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So y is equal to 2 when x is equal to 0.

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So c1-- when x is equal to 0, all the e terms

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you become 1, right?

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This one will become 1.

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And then notice, we have an xe to the 0.

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So now this x is 0.

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So this whole term is going to be equal to 0.

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So we're done. c1 is equal to 2.

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That was pretty straightforward.

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This x actually made it a lot easier.

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So c1 is equal to 2.

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And now we can use the derivative.

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So let's see, this is the first derivative.

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And I'll substitute c1 in there so we can

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just solve for c2.

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So our first derivative is y prime is equal to-- let's see

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c1-- 1/2 plus c2-- so it's-- well I'll write this first--

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it's equal to 2 over 2.

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So it's 1 plus c2 times e to the 1/2 x plus c2 over 2 times

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xe to the 1/2 x.

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There was an x here.

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So when x is equal to 0, y prime is equal to 1/3.

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So 1/3 is equal to-- well, x is equal to 0, this'll be 1--

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so it's equal to 1 plus c2.

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And then this term, when x is equal to 0, this whole thing

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becomes 0, right?

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Because this x just cancels out the whole thing.

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You multiply by 0 you get 0.

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So then we get 1/3 is equal to 1 plus c2, or that c2 is equal

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to 1/3 of minus 1 is equal to minus 2/3.

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And now we have our particular solution.

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Let me write it down and put a box around it.

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So this is our general solution.

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Our particular solution, given these initial conditions for

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this repeated root problem, is y is equal to c1-- we figured

144
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that out to be 2 fairly quickly-- 2e to the 1/2 x plus

145
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c2. c2 is minus 2/3.

146
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So minus 2/3 xe to the 1/2 x.

147
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And we are done.

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There is our particular solution.

149
00:08:05,579 --> 00:08:07,939
So once again, kind of the proof of how

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do you get to this.

151
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Why is there this x in there?

152
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And it wasn't a proof, it was really more of just to show

153
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you the intuition of where that came from.

154
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And it did introduce you to a method called, reduction of

155
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order, to figure out what that function v was, which ended up

156
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just being c1 plus c2 times x.

157
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But all that can be pretty complicated.

158
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But you see that once you know the pattern, or once you know

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that this is going to be the general solution, they're

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pretty easy to solve.

161
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Characteristic equation.

162
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Get your general solution.

163
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Figure out the derivative of the general solution.

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And then substitute your initial conditions to solve

165
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for your constants.

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And you're done.

167
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Anyway, I'll see you in the next video.

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And actually, we're going to start solving non-homogeneous

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differential equations.

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See

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