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Let's do another example of solving a nonhomogeneous

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linear differential equation with a constant coefficient.

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And the left-hand side is going to be the same one that

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we've been doing.

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The second derivative of y minus 3 times the first

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derivative minus 4 times y is equal to-- and now instead of

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having an exponential function or a trigonometric functional,

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we'll just have a simple-- well, it just looks an x

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squared term, but it's a polynomial.

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Right?

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And you know how to solve the general solution of the

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homogeneous equation if this were 0.

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So we're going to focus just now on the particular

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solution, then we can later add that to the general

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solution of a nonhomogeneous equation, to get the solution.

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So what's a good guess for a particular solution?

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Well, when we had exponentials, we guessed that

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our solution would be an exponential.

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When we had trigonometric functions, we guessed that our

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solution would be trigonomretric.

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So since we have a polynomial here that makes this

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differential equation nonhomogeneous, let's guess

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that a particular solution is a polynomial.

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And that makes sense.

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If you take a second-degree polynomial, take its

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derivatives and add and subtract, you should hopefully

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get another second-degree polynomial.

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So let's guess that it is Ax squared plus Bx plus C.

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And what would be a second derivative?

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Well a second derivative would be 2Ax plus B.

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Sorry, this is the first derivitive.

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The second derivative would be 2A.

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And now we can substitute back into the original equation.

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We get the second derivitive, 2A minus 3 times the first

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derivitive.

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So minus 3A-- oh no, sorry.

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Minus 3 times this.

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So minus 6Ax minus 3B minus 4 times the function itself.

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So minus 4Ax squared minus 4Bx minus 4C.

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That's just 4 times all of that.

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That's going to equal 4x squared.

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And I'll just group our x squared, our x and our

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constant terms, and then we could try to solve for the

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coeficients.

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So let's see.

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I have one x squared term here.

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So it's minus 4Ax squared.

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And then what are my x terms?

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I have minus 6Ax minus 4Bx.

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So then say plus minus 6A minus 4B times x.

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I just added the coefficients.

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And then finally we get our constant terms. 2A

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minus 3B minus 4C.

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And all of that will equal 4x squared.

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Now how do we solve for A, B, and C?

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Well, whatever the x squared coefficients add up on this

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side, it should equal 4.

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Whatever the x coefficients add up on this side, it should

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be equal to 0, right?

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Because you can view this as plus 0x, right?

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And then you could say plus 0 constant as well.

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So the constants should also add up to 0.

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So let's do that.

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So first let's do the x squared term.

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So minus 4A should be equal to 4.

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And then that tells us that A is equal to minus 1.

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Fair enough.

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Now the x terms. Minus 6A, minus 4B, that

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should be equal to 0.

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Right?

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So let's write that down.

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We know what A is, so let's substitute.

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So minus 6 times A, so minus 6 times minus 1.

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So that's 6 minus 4B is equal to 0.

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So we get 4B-- I'm just putting 4B on this side and

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then switching.

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4B is equal to 6.

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And B is equal to-- 6 divided by 4 is 3/2.

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And then finally the constant term should also equal 0, so

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let's solve for those.

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2 times A, that's minus 2.

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Minus 3 times B.

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Well, that's minus 3 times this.

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So minus 9/2 minus 4C is equal to 0.

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So let's see.

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I don't want to make a careless mistake.

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So this is minus 4 minus 9/2, right?

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That's minus 4/2 minus 9/2-- and we could take the 4C and

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put it on that side-- it's equal to 4C.

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What's minus 4 minus 9?

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That's minus 13/2.

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Minus 13/2 is equal to 4C.

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4C, divide both sides by 4, and then you get C is equal to

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minus 13/8.

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And I think I haven't made a careless mistake.

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So if I haven't, then our particular

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solution, we now know.

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Well, let me write the whole solution.

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So.

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And this is a nice stretch of horizontal real estate.

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So let's write our solution.

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Our solution is going to be equal to the particular

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solution, which is Ax squared, so that's minus 1x squared.

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Ax squared plus Bx plus 3/2x plus C minus 13/8.

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So this is the particular solution.

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We solved for A, B, and C.

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We determined the undetermined coefficient.

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And now if we want the general solution, we add to that the

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general solution of the homogeneous equation.

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What was that? y prime minus 3y prime minus

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4y is equal to 0.

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And we've solved this multiple times.

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We know that the general solution to the homogeneous

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equation is C1e to the 4x plus C2e to the minus x, right?

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You just take the characteristic equation r

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squared minus 3r minus 4.

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What did you get?

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You get r minus 4 times r plus 1, and then that's how you get

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minus 1 and 4.

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Anyway.

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So if this is the general solution to the homogeneous

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equation, this a particular solution to the

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nonhomogeneous equation.

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The general solution to the nonhomogeneous equation is

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going to be the sum of the two.

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So let's add that.

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So plus C1e to the 4x plus C2e to the minus x.

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So there you.

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I don't think that was too painful.

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The most painful part was just making sure that you don't

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make a careless mistake with the algebra.

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But using a fairly straightforward, really

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algebraic technique, we were able to get a fairly fancy

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solution to this second order linear nonhomogeneous

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differential equation with constant coefficients.

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See you in the next video.

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