I was given this problem for practice and was wondering if my approach was correct: $$ (1-x^2)y'' - 2xy' + 3ly = 0. $$
At first I thought of just using $l = 2$ since the Legendre DE is defined in that way (with the $l(l+1)y$ term). But I settled on first using the power series method to solve for the recurrence and then see if I could find $l$, but I'm completely stuck after that.
Any help would be greatly appreciated. Thanks!
I think the power series approach is the right one. If we assume that $$ y=\sum_{n=0}^ma_nx^n $$ with $a_m\neq0$, using the ODE yields $$ 0=(1-x^2)\sum_{n=0}^mn(n-1)a_nx^{n-2}-2x\sum_{n=0}^ma_nnx^{n-1}+3l\sum_{n=0}^ma_nnx^{n}\\ \Leftrightarrow 0=\sum_{n=0}^{m-2} \Big( (n+2)(n+1)a_{n+2} - n(n-1)a_n -2na_n+3la_n \Big)x^n+ \Big( - (m-1)(m-2) -2(m-1)+3l \Big)a_{m-1}x^{m-1} + \Big( - m(m-1) -2m+3l \Big)a_mx^m\\ \Leftrightarrow 0=\sum_{n=0}^{m-2} \Big( (n+2)(n+1)a_{n+2} - n(n+1)a_n+3la_n \Big)x^n+ \Big( - (m-1)m +3l \Big)a_{m-1}x^{m-1} + \Big( - m(m+1) +3l \Big)a_mx^m\\ $$ Since $a_m\neq0$, that means that $l=\frac{m(m+1)}{3}$ and $a_{m-1}=0$.