I have to derive corresponding differential equation for Legendre polynomial from Rodrigues formula: $$ P_n(x)=\frac{1}{2^n n!}\left(\frac{d}{dx}\right)^n(x^2-1)^n $$ The solution is indeed $(x^2-1)y''+2xy'-n(n+1)y=0$, but I am not sure, how to derive it.
My first idea was, to write something like $P_n''+A\cdot P_n'+B\cdot P_n=0$, but I wasn't sure, what is the derivative $P_n'$ (or $P_n''$)?
Any idea, how to proceed?
Let me work for simplicity here with the quantities $Q_n(x):=2^nn!P_n(x)$, i.e. $$ Q_n(x)=\partial_x^n((x^2-1)^n). $$
Compute the quantity $$ \partial_x^{n+1}((x^2-1)\partial_x((x^2-1)^n)) $$ in two different ways.
In one way we first perform the derivative inside: $$ \partial_x^{n+1}(2nx(x^2-1)^n)=2nx\partial_x^{n+1}((x^2-1)^n)+2n(n+1)\partial_x^{n}((x^2-1)^n)=2nxQ_n'(x)+2n(n+1)Q_n(x). $$
In the other way we first perform the derivative outside: $$ \partial_x^{n+1}((x^2-1)\partial_x((x^2-1)^n))= (x^2-1)\partial_x^{n+2}((x^2-1)^n)+ (n+1)\partial_x(x^2-1)\partial_x^{n+1}((x^2-1)^n)+ {n+1\choose 2}\partial_x^2(x^2-1)\partial_x^{n}((x^2-1)^n)= (x^2-1)Q_n''(x)+2(n+1)xQ'_n(x)+n(n+1)Q_n(x). $$
Equating the two expressions gives the equation $$ (x^2-1)Q_n''(x)+2xQ_n'(x)=n(n+1)Q_n(x). $$
