Given that $ P_l(x) $ is a polynomial solution of Legendre's equation, I want to find the differential equation for the function $ W(x) = (1-x^2) \frac{d^2}{dx^2}P_l(x)$
Since Legendre's equation is $(1-x^2)\frac{d^2}{dx^2}P_l(x) - 2x \frac{d}{dx}P_l(x) + l(l+1)P_l(x)=0 $ I can right away write that $$ W(x) = 2xP'_l(x) - l(l+1)P_l(x) $$ But then, I don't know how to write this as a differential equation for W, since it seems like I only know W in terms of integrals of itself. Thanks.
We have $$W(x) = 2 x P_l'(x) - l(l+1) P_l(x) $$ Differentiating twice and using the differential equation to express the results in terms of $P_l$ and $P_l'$, look at $W'' + a W' + b W$, and solve for $a$ and $b$ to make this $0$: I get $a = \dfrac{2x}{x^2-1}$, $b = -\dfrac{l^2+l}{x^2-1} - \dfrac{4}{(x^2-1)^2}$, i.e. $$ W'' + \dfrac{2x}{x^2-1} W' - \left(\dfrac{l^2+l}{x^2-1} + \dfrac{4}{(x^2-1)^2}\right) W = 0 $$