Relationship between Legendre polynomials and Legendre functions of the second kind I'm taking an ODE course at the moment, and my instructor gave us the following problem:

Derive the following formula for Legendre functions $Q_n(x)$ of the second kind:
$$Q_n(x) = P_n(x) \int \frac{1}{[P_n(x)]^2 (1-x^2)}dx$$
where $P_n(x)$ is the $n$-th Legendre polynomial.

He introduced Legendre functions in the context of second order ODEs,  but we haven't really used them for anything - moreover, this is the only problem we were assigned that has anything to do with them. As a result, I'm sort of at a loss of where to start.
I've tried a couple of things (like using the actual Legendre ODE
$$(1-x^2)y^{\prime \prime} - 2xy^{\prime} + n(n+1)y = 0$$
and plugging in the solution $y(x)=a_1P_n(x)+a_2Q_n(x)$ and proceeding from there) but so far, haven't been able to go anywhere.
Any help (preferably as elementary as possible) would be much appreciated. Thanks!

The most general solution of Legendre equation is
$$y = A{P_n} + B{Q_n}.$$
Let $y(x) = A(x){P_n}(x)$. Then $y' = AP' + A'P$ and $y'' = AP'' + 2A'P' + A''P$. So
$$(1 - {x^2})(AP'' + 2A'P' + A''P) - 2x(AP' + A'P) + n(n + 1)AP = 0.$$
Note that
$$(1 - {x^2})(AP'') - 2x(AP') + n(n + 1)AP = 0$$
which means some terms in the above equation vanish.
Now let $A' = u$ and reduce the order so that
$$2\frac{{dP}}{P} + \frac{{du}}{u} - \frac{{2xdx}}{{1 - {x^2}}} = 0$$
and
$$u = \frac{{{\text{const}}}}{{(1 - {x^2}){P^2}}}$$
so
$$A = {C_n}\int {\frac{1}{{(1 - {x^2}){P^2}}}dx}.$$
See if you can do the rest.
