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?

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 $A=\frac{x}6$.
