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Before we move on past the method of undetermined

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coefficients, I want to make and interesting and actually a

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useful point.

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Let's say that I had the following nonhomogeneous

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differential equation: the second derivative of y minus 3

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times the first derivative minus 4y is equal to-- now

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this is where gets interesting-- 3e to the 2x

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plus 2 sine of x plus-- let me make sure that I'm doing the

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same problems that I've already worked

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on-- plus 4x squared.

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So you might say, wow, this is a

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tremendously complicated problem.

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I have the 3 types of functions I've been exposed

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to, I would have so many undetermined coefficients, it

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would get really unwieldy.

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And this is where you need to make a simplifying

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realization.

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We know the three particular solutions to the following

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differential equations.

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We know the solution to the second derivative minus 3

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times the first derivative minus 4y.

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Well, this is this the homogeneous, right?

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And we know that the solution to the homogeneous equation--

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we did this a bunch of times-- is C1e to the 4x plus C2e to

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the minus x.

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We know the solution to-- and I'll switch colors, just for a

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variety-- y prime prime minus 3y prime minus 4y is equal to

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just this alone: 3e to the 2x.

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And we saw that that particular solution there, y

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particular, was minus 1/2 e to the 2x.

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And we did this using undetermined coefficients.

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We did that a couple of videos ago.

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And then let me just write this out a couple of times.

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We know the solution to this one, as well.

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This was another particular solution we found.

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I think it was two videos ago.

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And we found that the particular solution in this

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case-- and this was a fairly hairy problem-- was minus 5/17

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x plus 3/17.

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Sorry.

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The particular solution was minus 5/17 sine of x plus 3/17

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cosine of x.

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And then finally this last polynomial, we could call it.

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We know the solution when that was just the right-hand side.

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That was this equation.

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And there we figured out-- and this was in the last video.

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We figured out that the particular solution in this

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case was minus x squared plus 3/2 x minus 13/8.

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So we know the particular solution when 0's on the

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right-hand side.

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We know it when just 3e to the 2x is on the right-hand side.

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We know it just when 2 sine of x is on the right-hand side.

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And we know it just when 4x squared is on

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the right-hand side.

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First of all, the particular solution to this

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nonhomogeneous equation, we could just take the sum of the

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three particular solutions.

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And that makes sense, right?

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Because one of the particular solutions, like this one when

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you put it on the left-hand side, it will

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just equal this term.

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This particular solution, when you put it in the left-hand

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side, will equal this term.

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And finally, this particular solution, when you put it on

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the left-hand side, will equal the 4x squared.

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And then you could add the homogeneous solution to that.

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You put it in this side and you'll get 0.

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So it won't change the right-hand side.

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And then you will have the most general solution because

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you have these two constants that you can solve for

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depending on your initial condition.

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So the solution to this seemingly hairy differential

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equation is really just the sum of these four solutions.

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Let me clean up some space because I want everything to

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be on the board at the same time.

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So the solution is going to be-- well, I

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want that to be deleted.

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I'll do it in baby blue.

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Is going to be the solution to the homogeneous C1e to the 4x

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plus C2e to the minus x minus 1/2 e the to 2x.

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And I'll continue this line.

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Minus 5/17 sine of x plus 3/17 cosine of x minus x squared

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plus 3/2 x minus 13/8.

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And it seems daunting.

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When you saw this, it probably looked daunting.

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This solution, if I told you this was a solution and you

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didn't know how to do undetermined coefficients,

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you're like, oh, I would never be able to figure out

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something like that.

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But the important realization is that you just have to find

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the particular solutions for each of these terms and then

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sum them up.

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And then add them to the general solution for the

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homogeneous equation, if this was a 0 on

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the right-hand side.

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And then you get the general solution for this fairly

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intimidating-looking second order linear nonhomogeneous

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differential equation with constant coefficients.

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See you in the next video, where we'll start learning

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another method for solving nonhomogeneous equations.

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