In an application I encountered the ODE $$ \left( {x}^{2}-1 \right) {\frac {{\rm d}^{2}}{{\rm d}{x}^{2}}}f \left( x \right) +x \left( {\frac {\rm d}{{\rm d}x}}f \left( x \right) \right) \left( 8\,{x}^{2}-7 \right) -4\, \left( C+1 \right) f \left( x \right) =0. $$ In principle, this is Legendre's differential equation with an additional term $\tilde \quad x^3 \frac{d}{dx}f(x)$ and with a solution that can be expressed by Maple as $f(x) = K_1 HeunC(4, -\frac{1}{2}, -\frac{1}{2}, -2, \frac{3}{8}-C, x^2)$, where $K_1 \in \mathbb{R}$ satisfying the boundary conditions indicated below. Of course, this confluent Heun's function does not tell me anything if I am looking for actual solutions of this Sturm-Liouville problem that can be expressed in terms of analytical functions( which may or may not exist). The main problem here is that the unknown eigenvalues are part of this solution. Especially, it does not help me identifying the eigenvalues ( which should behave asymptotically Schrödinger-operator like $C_n \tilde \quad n^2$). So I was only able to identify one 'easy solution' $f(x) = 1$ and $C=-1$. But this is it. I don't see how I can find any other solution analytically. Does anybody here know whether solutions to this function are known or whether there are any solutions that we can write down? Has this particular ODE ever been studied? I am greatful to every comment.

Edit: I am looking for bounded solutions (eigenfunctions and eigenvalues) on $[-1,1]$ satisfying $f(-1) = f(1)$ and $f'(-1) = f'(1)$. I am interested in finding explicit representations (sure this is a vague term, but probably you know what I am talking about. A function defined by a series or an integral is an explicit representation, but a confluent heun's function that is just defined as a solution to a ODE is not). I am aware of the fact that there is a high chance that things like recurrence relations cannot be found. Despite, you may be able to identify how solutions could look like (e.g. a polynomial of degree n). This would also help very much.