$\frac{d^2 y}{dx^2}-2y=2\tan^3\left(x\right)$ Problem:
\begin{equation}
\frac{d^2 y}{dx^2}-2y=2\tan^3\left(x\right).
\end{equation}
using the method of undetermined coefficients or variation of parameters, with $y_p\left(x\right)=\tan\left(x\right)$.

This is what I have so far:
\begin{equation}
r^2-2=0\implies r=\pm\sqrt{2},\:\:\:\:\:\therefore\:\: C_1 e^{\sqrt{2}t}+C_2 e^{-\sqrt{2}t},
\end{equation}
so that is my "complimentary solution" but I do not know what to do with it.

I don't have the reputation to comment, but the general solution of the nonhomogeneous equation is just the fundamental set of solutions to the homogeneous equation added to the particular solution. So you have everything, just put it all together.
