Solving a differential equation with the method of series as described in this site, I arrived to this point:
$$\sum_nr^n\left[a_{n-1}(2n+n^2-3+4\rho+4n\rho+4\rho^2)+a_n(l(l+1)-2n+3-4\rho)+a_{n+1}(-l(l+1)-2n\rho-4\rho^2)\right]=0$$
how do I find the general term of the series?
If there where only two terms, I would be able to solve it as described in the above site, but there are three terms and I do not know how to proceed. For example if the series was of the type:
$$\sum_n r^n(A a_n+ B a_{n+2})$$ I would divide by A and write a table with two columns for $n$ and for $n+2$, so I would find the recursion relation, but here there is an additional term and I do not know how to consider it to solve the recurrence relation.
The differential equation is:
$$(r-1)rf^{\prime\prime}(r)+f^\prime(r)-\left[ \dfrac{\rho^2 r^3}{r-1}+l(l+1)-\dfrac{3}{r} \right]f(r)=0$$