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.