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I'm having a little trouble understanding how to solve the following differential equation. The equation that has to be solved is: $$y'' - y' - 2y = 2e^{-t}$$

The problem I am having is that I don't understand why they equate that part with the derivatives of the $u$ parameters to $0$. See the problem below

Example 13. Problem 2: $y''-y'-2y=2e^{-t}$. The characteristic equation is: $$r^2-r-2=0\iff (r-2)(r-1)=0\implies y_h(t)=c_1e^{2t}+c2e^{-t}.$$ Suppose that $y(t)=u_1(t)e^{2t}+u_2(t)e^{-t}$, then it follows: $$ y'(t)=\underbrace{u_1'(t)e^{2t}+u_2'(t)e^{-t}}_{=0}+2u_1(t)e^{2t}-u_2(t)e^{-t} $$

Here they first find the characteristic equation and write down the general solution. They then replace the constants with the parameter "$u$" and take the derivative.

As you can see, they just say that the derivative part of the $u$ parameter is equal to $0$. But why? How? Where did that come from? I can't find it anywhere in my book.

It's probably a facepalm answer but I would really appreciate it.

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I kind of get that since e^2t and e^-t are solutions, any linear combination of those two is equal to zero. But how come it's only the case when it's u' and not just with u? How can we know that u' is a constant and not u? This is what confuses me –  Ortix92 Jan 7 '13 at 14:56
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1 Answer

up vote 2 down vote accepted

Read through the derivation of the method of variation of parameters here and it should clarify things. The very method is predicated on finding $u_1(x)$, $u_2(x)$ such that $y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$ is a solution of the nonhomogeneous problem, where $y_1(x)$, $y_2(x)$ are known (linearly independent) solutions of the homogeneous problem.

So we are looking for the yet-to-be-determined functions $u_1$ and $u_2$. While doing so, we can employ any constraints we like; such constraint(s) will be fruitful so long as they ultimately accomplish our goal of finding $u_1$, $u_2$ for which $y_p$ above is a solution to the nonhomogeneous problem.

Now your question may really be, "How did they think to do this? Because I wouldn't have thought to do that." Fair enough question/comment. However, now that you have seen this "trick" and that it "worked" in the sense of eventually finding you a second solution, just rejoice and use the trick again (while hopefully understanding that it was mathematically legitimate to impose that constraint to make the second derivative simpler).

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Thanks for your help! It was indeed never fully explained, but it did become a lot clearer now. I guess it's just a matter of practice to "see" whenever to apply this "trick" –  Ortix92 Jan 7 '13 at 15:27
    
Well, now that you have seen it once, just remember that you will make exactly the same constraint every time you employ the method of variation of parameters. Basically when you get to the underbraced step in the work in your post, just say to yourself, "I am imposing this constraint on $u_1$, $u_2$ and will see if it bears fruit." It always will. ;-) –  JohnD Jan 7 '13 at 15:30
    
To the point of being dumb and repetitive, may I ask why this constraint always bears fruit? –  Jayesh Badwaik Oct 8 '13 at 16:09
    
@JayeshBadwaik: Not being coy, but it simply works (at least for that particular constraint), because it works, i.e. once that constraint is imposed, the leftover terms (see the linked article) are such that one can actually get a solution solve for the unknown function. Of course one is not guaranteed of similar success for any constraint, but it is at least successful for that constraint. –  JohnD Oct 9 '13 at 0:28
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