Finding particular solution to $y'' + 2y' - 8y = e^{2x} $ $$y'' + 2y' - 8y = e^{2x} $$
How do I find the particular solution?
I tried setting: $y = Ae^{2x} => y' = 2Ae^{2x} => y''= 4Ae^{2x}$
If I substitute I get: $4Ae^{2x} + 4Ae^{2x} - 8Ae^{2x} = e^{2x} => 0 = e^{2x}$
What am I doing wrong?
 A: That is because $e^{2x}$ is a solution of the homogeneous equation. You can see this from solving the characteristic equation:
$$r^2+2r-8=0$$
It gives you two solutions $-4$ and $2$, which means $e^{2x}$ is a solution to the homogeneous equation. 
So for the particular solution, try $Axe^{2x}$ instead. If it fails, try $Axe^{2x}+Bx^2e^{2x}$, and so on.
A: You already received answers.
When you have a doubt (and when your work leads to something as $0=e^{2x}$ as you honestly pointed out), just do what you did but considering now that $A$ is a function of $x$. So, $$y=A\,e^{2x}$$ $$y'=A'\,e^{2x}+2A\,e^{2x}$$ $$y''=A''\,e^{2x}+4A'\,e^{2x}+4A\,e^{2x}$$ Plug all of that in the differential equation, divide both sides by $e^{2x}$ and get $$A''+6A'=1$$ So $A$ cannot be a constant. The simplest should be $A=\frac{x}6$.
A: The particular solution of 
\begin{equation*}
y''-8y+2y=e^{2x}
\end{equation*}
is going to be of the form $y_p=x(c_1e^{2x})$ (the multiplication by $x$ is to take account for the fact that $e^{2x}$ is in the complementary solution). Then
\begin{equation*}
y'_p=(c_1e^{2x}x)'\\
\Rightarrow y''_p=c_1(4e^{2x}+4e^{2x}x).
\end{equation*}
Now substitute $y_p$ back into the differential equation and equate coefficients to figure out $c_1.$
