Solving $y''(x) - [B + A \sin(x)]y(x) = 0$ for periodic $y$

Question

Does anybody know how to solve this differential equation: $$ y''(x) - \left[B + A \sin(x)\right]y(x) = 0 $$ if $y$ is periodic: $y(x + 2\pi) = y(x)$. I have tried looking for solution in form of Fourier series: $$ y = \sum_n a_n\exp(inx) $$ and arrived to recursive equation for Fourier coefficients: $$ (B - n^2)a_n = A'[a_{n + 1} - a_{n -1}], \quad \mbox{where } A' = \frac{A}{2i}. $$ Don't know what to do next.

Answer 1

Aside from minor changes of variables this is Mathieu's differential equation $$w''(z) + \left[a - 2q \cos(2z)\right]w(z) = 0$$ (Specifically, one needs $w=y,z=\pi/4-x/2,q=2A,a=-4B$.) What you specifically want are solutions with $w(z+\pi)=w(z)$, corresponding to two of the four eigenfunctions listed in section 28.2(vi) of the DLMF. (Two more boundary conditions are needed to determine a particular linear combination of these eigenfunctions.)

It must be stressed, though that these are transcendental functions of $x$ and can't be solved in closed form; the same goes for the Fourier coefficients, whose recurrence relation can't be solved in closed-form. So one must either rely on numerical methods or asymptotic behavior, the latter of which is also documented at DLMF.