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?

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 $c_1.$
