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Let's do some more nonhomogeneous equations.

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So let's take the same problem, but we'll change the

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right-hand side.

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Because I think you know how to solve the-- essentially,

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the homogeneous version.

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So the same problem as we did in the last video.

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The second derivative of y minus 3 times the first

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derivative y minus 4 times the function.

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And now in the last example, the nonhomogeneous part

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was 3e to the 2x.

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But we're tired of dealing with exponent functions, so

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let's make it a trigonometric function.

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So let's say it equals 2 sin of x.

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So the first step you do is what we've been doing.

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You essentially solve the homogeneous equation.

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So this left-hand side is equal to 0.

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You do that by getting the characteristic equation r

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squared minus 3r minus 4 is equal to 0.

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You get the solutions, r is equal equal to 4, r is equal

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to minus 1, and then you get that general solution.

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We did this in the last video.

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You get the general solution of the homogeneous.

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Maybe we'll call this the homogeneous solution.

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y homogeneous.

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We've got the C1 e to the 4x plus C2e to the minus x.

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And that's all and good, but in order to get the general

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solution of this nonhomogeneous equation, I

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have to take the solution of the nonhomogeneous equation,

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if this were equal to 0, and then add that to a particular

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solution that satisfies this equation.

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That satisfies-- when you take the second derivative minus 3

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times the first minus 4 times the function, I actually

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get 2 sin of x.

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And here once again we'll use undetermined coefficients.

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And undetermined coefficients, just think to yourself.

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What function, when I take its second and first derivatives

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and add and subtract multiples of them to each other, will I

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get sine of x?

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Well, two functions end up with sine of x when you take

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the first and second derivatives.

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And the sine and cosine of x.

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So it's a good guess.

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And that's really what you're doing it the method of

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undetermined coefficients.

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You take a guess of a particular solution and then

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you solve for the undetermined coefficients.

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So let's say that our guess is y is equal to-- I don't know,

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some coefficient times sine of x.

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And if this was sine of 2x, I'd put A

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times sine of 2x here.

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Just because I still want-- no matter what happens here-- the

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sine of 2x's or maybe cosine of 2x's to still exist. If

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this was a sine of 2x, there's nothing I could do to a sine

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of x, or nothing at least trivial that I could do

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to the sine of x.

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It would end up with a sine of 2x.

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So whatever's here, I want here.

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Plus B, some undetermined coefficient times cosine of x.

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And once again, this was sine of 2x.

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I'd want a cosine of 2x here.

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So let's figure out its first and second derivitives.

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So the first derivative of this y prime is equal to A

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cosine of x.

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Cosine derivative is minus sine, so minus B sine of x.

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And then the second derivitive--

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I'll write down here.

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The second derivative is equal to what?

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Derivative of cosine is minus sine, so minus A sine of x

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minus B cosine of x.

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I think you're starting to see that the hardest thing in most

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differential equations problems is not making

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careless mistakes.

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It's a lot of algebra and a lot of fairly basic calculus.

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And the real trick is to not make careless mistakes.

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Every time I say that, I tend to make one.

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So I'm going to focus extra right now.

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So anyway, let's take these and substitute them back into

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this nonhomogeneous equation.

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Let's see if I can solve for A and B.

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So the second derivative is that.

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Let me just rewrite it, just so that you

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see what I'm doing.

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So I'm going to take the second derivative, y prime

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prime, so that's minus A sine of x minus B cosine of x.

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I'm going to add minus 3 times the first derivative to that.

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And I'm going to write the sines under the sines and the

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cosines under the consines.

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So minus 3 times this.

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So the sine is, let's see.

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It's plus 3B sine of x minus 3 times this.

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So minus 3A cosine of x.

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And then minus 4 times our original function.

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So minus 4A sine of x.

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Right?

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Minus 4 times that.

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Minus 4 times this.

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Minus 4B cosine of x.

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And when I take the sum of all of those-- that's essentially

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the left-hand side to this equation-- when I take the sum

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of all of that, that is equal to 2 sine of x.

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I could have written them out in a line, but it would have

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just been more confusing.

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And now this makes it easy to add up the sine of x's and the

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cosine of x's.

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So if I add up all the coefficients on the sine of x,

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I get minus A plus 3B minus 4A.

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So that looks like minus 5A plus 3B sine of x plus-- and

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now what are the coefficients here?

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I have minus B and then I have another minus 4B, so minus 5B

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and then minus 3A.

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So minus 3A minus 5B cosine of x.

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The cosine of x should go right here.

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So anyway, how do I solve for A and B?

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Well, I have the minus 5A 3B is equal to whatever

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coefficients in front of sine of x here.

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So minus 5A plus 3B must be equal to 2.

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And then minus 3A minus 5B is the coefficient on cosine of

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x, although I kind of squeezed in the

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cosine of x here, right?

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So this must be equal to whatever the coefficient on

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cosine of x is on the right-hand side.

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Well the coefficient of cosine of x on the

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right-hand side is 0.

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So that sets up a system of two

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unknowns with two equations.

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A linear system.

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So we get minus 5A plus 3B is equal to 2.

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And we get minus 3A minus 5B is equal to 0.

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And let's see if I can simplify this a little bit.

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Let's see.

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This is a system of two unknowns, two equations.

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If I multiply the top equation by 5 1/3s, right?

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Actually, let me multiply the top equation by 5 1/3s.

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I get minus 25/3 A plus 5B is equal to 5 1/3s times this.

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5 1/3s times 2 is 10 1/3s.

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And the bottom equation is minus 3A minus

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5B is equal to 0.

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Let's add the two equations.

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I get 10 1/3s is equal to-- these cancel out.

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That's minus 25/3 minus 9/3 A is equal to 10 1/3s.

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This is getting a little bit messier than I like, but we'll

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soldier on.

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So minus 25 minus 9.

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What's minus 25 minus 9?

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So that is 34.

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So we get 34 over 3A is equal to 10/3.

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We can multiply both sides by 3.

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Divide both sides by 34.

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A is equal to 10/34, which is equal to 5/17.

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Nice ugly number.

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5/17 and now we can solve for B.

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So let's see.

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Minus 3 times A minus 3 times A.

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5/17 minus 5B is equal to 0.

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So that's what?

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Minus 15/17 is equal to plus 5B.

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I just took this and put it on the right-hand side.

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And then divide both sides by 5.

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Oh, you know what?

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I realized I made a careless mistake here.

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Minus 25 minus 9.

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That's the minus 34 over 3.

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so minus 34A is equal to 10.

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A is equal to minus 10/34 or minus 5/17.

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So minus 3 times minus 5/17.

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So 5/17 is equal to plus 5B, right?

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And then we get B is equal to 3/17.

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That was hairy.

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And notice, the hard part was not losing

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your negative sines.

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But anyway, we now have our particular solution to this.

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let me try to write in a non-nauseating color, although

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I think I picked a nauseating one.

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The particular solution is A minus 5/17 sine of x-- right?

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That was a coefficient on sine of x-- plus B plus 3/17 times

183
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cosine of x.

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And if we look at our original problem, the general solution

185
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out of this nonhomogeneous equation would be this-- which

186
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is the general solution to the homogeneous equation, which

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we've done many videos on-- plus now our particular

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solution that we solved using the method of undetermined

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coefficient.

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So if you just take that and add it to that, you're done.

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And I am out of time.

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See you in the next video.