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We're now ready to solve non-homogeneous second-order

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linear differential equations with constant coefficients.

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So what does all that mean?

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Well, it means an equation that looks like this.

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A times the second derivative plus B times the first

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derivative plus C times the function is equal to g of x.

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Before I show you an actual example, I want to show you

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something interesting.

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That the general solution of this non-homogeneous equation

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is actually the general solution of the homogeneous

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equation plus a particular solution.

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I'll explain what that means in a second.

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So let's say that h is a solution of

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the homogeneous equation.

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And that worked out well, because, h for homogeneous.

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h is solution for homogeneous.

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There should be some shorthand notation for homogeneous.

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So what does that mean?

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That means that A times the second derivative of h plus B

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times h prime plus C times h is equal to 0.

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That's what I mean when I say that h is a solution-- and

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actually, let's just say that h is the general solution for

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this homogeneous equation.

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And we know how to solve that.

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Take the characteristic equation depending on how many

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roots it has and whether they're real or complex.

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You can figure out a general solution.

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And then if you have initial conditions, you can substitute

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them and get the values of the constants.

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Fair enough.

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Now let's say that I were to say that g is a solution.

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Well no, I already used g up here.

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Well, I don't like using vowels.

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Let's say j.

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Let's say j is a particular solution to this

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differential equation.

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So what does that mean?

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That means that A times j prime prime plus B times j

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prime plus C times j is equal to g of x.

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Right?

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So we're just defining j of x to be a particular solution.

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Now what I want to show you is that j of x plus h of x is

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also going to be a solution to this original equation.

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And that it's the general solution for this

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non-homogeneous equation.

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And before I just do it

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mathematically, what's the intuition?

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Well, when you substitute h here, you get 0.

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When you substitute j here, you get g of x.

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So when you add them together, you're going to get

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0 plus g of x here.

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So you're going to get g of x here.

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And I'll show you that right now.

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So let's say I wanted to substitute h plus j here.

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And I'll do it in a different color.

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A-- so the second derivative of the sum of those two

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functions is going to be the second derivative of both of

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them summed up-- plus B times the first derivative of the

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sum plus C times the sum of the functions.

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And my goal is to show that this is equal to g of x.

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So what is this simplified to?

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Well if we take all the h terms, we get Ah prime prime

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plus Bh prime plus Ch plus, let's do all the j terms. Aj

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prime prime plus Bj prime plus Cj.

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Well by definition of how we defined h and j, what

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is this equal to?

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We said that h is a solution for the homogeneous equation,

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or that this expression is equal to 0.

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So that equals 0.

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And by our definition for j, what does this equal?

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We said j is a particular solution for the

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non-homogeneous equation, or that this expression is

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equal to g of x.

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So when you substitute h plus j into this differential

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equation on the left-hand side.

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On the right-hand side, true enough, you get g of x.

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So we've just shown that if you define h and j this way,

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that the function, we'll call it k of x is equal to h of x

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plus j of x.

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I'm running out of space.

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That is the general solution.

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I haven't proven that is the most general solution, but I

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think you have the intuition, right?

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Because the general solution on the homogeneous one that

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was the most general solution, and now we're adding a

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particular solution that gets you the g of x on the

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right-hand side.

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That might be very confusing to you, so let's actually try

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to do it with some real numbers.

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And I think it'll make a lot more sense.

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Let's say we have the differential equations-- and

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I'm going to teach you a technique now for figuring out

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that j in that last example.

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So how do you figure out that particular solution?

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Let's say I have the differential equation the

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second derivative of y minus 3 times the first derivative

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minus 4 times y is equal to 3e to the 2x.

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So, the first step is we want the general solution of the

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homogeneous equation.

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And in that example I just did, that would have

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been our h of x.

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So we want the solution of y prime prime minus 3y prime

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minus 4y is equal to 0.

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Take the characteristic equation.

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This 4 is equal to 0.

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r minus 4 times r plus 1 is equal to 0.

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2 roots, r could be 4 or negative 1.

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And so our general solution-- I'll call that h.

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Well, let's call that y general.

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y sub g.

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So our general solution is equal to-- and we've done this

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many times-- C1 e to the 4x plus C2 e to the minus

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1x, or minus x.

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Fair enough.

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So we solved the homogeneous equation.

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So how do we get, in that last example, a j of x that will

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give us a particular solution, so on the

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right-hand side we get this.

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Well here we just have to think a little bit.

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And this method is called The Method of Undetermined

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Coefficients.

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And you have to say, well, if I want some function where I

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take a second derivative and add that or subtracted some

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multiple of its first derivative minus some multiple

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of the function, I get e to the 2x.

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That function and its derivatives and its second

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derivatives must be something of the form, something

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times e to the 2x.

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So essentially we take a guess.

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We say well what does it look like when we take the various

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derivatives and the functions and we multiply multiples of

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it plus each other?

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And all of that.

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We would get e the to 2x or some multiple of e to the 2x.

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Well, a good guess could just be that j-- well I'll call it

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y particular.

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Our particular solution here could be that-- and particular

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solution I'm using a little different than the particular

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solution when we had initial conditions.

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Here we can view this as a particular solution.

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A solution that gives us this on the right-hand side.

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So let's say that the one I pick is some constant A times

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e to the 2x.

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If that's my guess, then the derivative of that is equal to

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2Ae the to 2x.

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And the second derivative of that, of my particular

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solution, is equal to 4Ae to the 2x.

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And now I can substitute in here, and let's see if I can

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solve for A, and then I'll have my particular solution.

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So the second derivitive, that's this.

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So I get 4Ae to the 2x minus 3 times the first derivitive.

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So minus 3 times this.

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So that's minus 6Ae to the 2x minus 4 times the function.

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So minus 4Ae to the 2x, and all of that is going to be

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equal to 3e to the 2x.

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Well we know e to the 2x equal 0, so we can divide

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both sides by that.

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Just factor it out, really.

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Get rid of all of the e's to the 2x.

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On the left-hand side, we have 4A and a minus 4A.

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Well, those cancel out.

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And then lo and behold, we have minus 6A is equal to 3.

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Divide both sides by 6 and get A is equal to minus 1/2.

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So there.

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We have our particular solution.

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It is equal to minus 1/2 e to the 2x.

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And now, like I just showed you before I cleared the

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screen, our general solution of this non-homogeneous

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equation is going to be our particular solution plus the

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general solution to the homogeneous equation.

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So we can call this the most general

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solution-- I don't know.

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I'll just call it y.

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It is our general solution C1e to the 4x plus C2e to the

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minus x plus our particular solution we found.

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So that's minus 1/2e to the 2x.

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Pretty neat.

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Anyway, I'll do a couple more examples of this.

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And I think you'll get the hang of it.

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In the next examples, we'll do something other than an e to

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the 2x or an e function here.

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We'll try to do stuff with polynomials and trig

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functions as well.

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I'll see you in the next video.

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