The differential equation that is satisfied by the Legendre's polynomials is:
$$(1-x^2)y'' - 2xy' + \lambda y = 0 (*)$$
I have also been told that the Legendre's polynomial with the parameter $x = \cos\theta$ satisfies the spherical harmonics:
$$\sin\theta \frac{d}{d\theta}(\sin \theta \frac{dy}{d\theta}) - (m^2 - \lambda\sin^2(\theta))y = 0$$
which is equivalent to (under the substitution $x = \cos \theta$):
$$(1-x^2)y'' - 2xy' - (\frac{m^2}{1-x^2}- \lambda)y = 0 (**)$$
and the solution for this differential equation is $P(\cos \theta)$. However, I don't see how $(*)$ and $(**)$ compare? The first scalar term in $(*)$ is a constant $\lambda$, but the scalar term in $(**)$ is not: $\frac{m^2}{1-x^2} - \lambda$. Why is are lagrange polynomials a solution for this equation?
The solutions to (**) (when $\lambda=\ell(\ell+1)$ and $0\leq m\leq \ell$, with $\ell,m\in\mathbb{N}$) are the associated Legendre polynomials, of which the Legendre polynomials are a special case, corresponding to $m=0$.