solution of second order ODE with non-constant coefficients What is the general solution of $\frac{d^2y}{dx^2}+P[Q-R\cosh(Sx)]y=0$ where $P,Q,R,S$ are real and positive? I tried transforms but cannot get a solution.

$$\frac{d^2y}{dx^2}+P[Q-R\cosh(Sx)]y=0$$
This ODE is of the Mathieu's ODE kind : $\frac{d^2y}{dz^2}+\left(a-2q\cos(2z) \right)=0$ which general solution is expressed with the Mathieu's functions :
$$y=c_1 \text{MathieuC}(a,q,z)+c_2 \text{MathieuS}(a,q,z)$$
http://mathworld.wolfram.com/MathieuFunction.html
With $\begin{cases}
a=-\frac{4PQ}{S^2}\\ q=-\frac{2PR}{S^2} \\ z=\frac{iSx}{2} \end{cases}$
$$ y=c_1 \text{MathieuC}\left(-\frac{4PQ}{S^2}\:,\:-\frac{2PR}{S^2}\:,\:\frac{iS}{2}x\right)+c_2 \text{MathieuS}\left(-\frac{4PQ}{S^2}\:,\:-\frac{2PR}{S^2}\:,\:\frac{iS}{2}x\right)$$
This is a purely formal result. In practice, better use numerical calculus to solve the ODE.
