In my work, I derived an equation that looks like this:
\begin{align} y''+(A+\epsilon\sin\Omega t+B\sin\omega t)y=C\cos\omega t \end{align} where $\epsilon\ll1$. This is similar to the Whittaker-Hill differential equation, except $\omega$ may not be equal to $2\Omega$. I am wondering the best way to go about solving this. I have two ideas.
Idea 1
As $\epsilon<<1$, we can solve the equation perturbatively. Therefore, if we let $y=y_0+\epsilon y_1$, then we find the differential equations of
\begin{align} y_0''+(A+B\sin\omega t)y_0=C\cos\omega t \end{align} \begin{align} y_1''+(A+B\sin\omega t)y_1=-y_0\sin\Omega t \end{align}
These look like Mathieu differential equations with a forcing term. However, I'm not sure how to solve the resulting equations. I tried doing a literature search, but the generalizations of the Mathieu differential equation I found tended to be on including non-linear terms, such as in here and here. I have also found a paper considering $\epsilon=B$, but in my case, this is not necessarily true.
Idea 2
As the top equation looks like the Whittaker-Hill differential equation, then perhaps there is a generalization of it that I can use. However, I have not found anything useful.
Any insights would be greatly appreciated. Thanks.