ODE with additional term In an application I encountered the ODE 
$$ \left( x^2-1 \right) \frac {{\rm d}^{2}}{{\rm d} x^2} f
 ( x ) +x \left( \frac {\rm d}{{\rm d}x} f (x)  \right) ( 8x^2-7 ) -4 (C+1) f( x ) =0.
$$
 I don't see how I can find any 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. In particular, just a few more solution are helpful too, if you cannot identify them all.
 A: You should be able to apply the Frobenius method here.  Writing $f(x)=\sum_{n=0}^{\infty}a_n x^n$ and substituting into your differential equation gives the following:
$$
(x^2-1)\sum_{n=0}^{\infty}n(n-1)a_n x^{n-2} + x(8x^2-7)\sum_{n=0}^{\infty}na_n x^{n-1} - 4(C+1)\sum_{n=0}^{\infty}a_n x^n=0.
$$
The coefficient of $x^n$ is
$$
\left(-n(n+6)-4(C+1)\right)a_n +(n+1)(n+2)a_{n+2}+8(n-2)a_{n-2},
$$
where the $a_{n-2}$ term only appears for $n\ge 3$.  Since this coefficient must vanish for each $n$, you have
$$
a_2 = 2(C+1)a_0
$$
and
$$
a_{n+2}=\frac{\left(n(n+6)+4(C+1)\right)a_n - 8(n-2)a_{n-2}}{(n+1)(n+2)}
$$
for $n\ge 2$ in the even solution (which is the one you're interested in).  Now, unlike the Legendre differential equation, this recursion will not terminate for any obvious choice of $C$ (other than $C=-1$, as you pointed out).  But we can look, numerically at least, for choices where the terms go to zero.  (If they instead go to a constant, then the power series diverges at $x=1$.)  Doing this gives eigenvalues at approximately $C=1, -1, -2.75, -6.59, -12.55, -20.53, \ldots$.  The eigenvalues converge rapidly to the usual spectrum, $C_l=-l(l+1)-1/2$ for $l\in \mathbb{N}$.
