Legendre Diff. Eq. Appearing in Polar Equation of Hydrogen Atom

Question

The usual form of Legendre's differential equation which I am familiar with, is: $$ \left(1-x^2\right)\frac{\mathrm d^2P}{\mathrm dx^2} - 2x\frac{\mathrm dP}{\mathrm dx} + \ell\left(\ell+1\right)P = 0 \tag{01} $$

No problem, use series solution to solve.

But when I was looking at the solution of Schrodinger's equation for Hydrogen atom, the equation they got was this: $$ \left(1-x^2\right)\frac{\mathrm d^2P}{\mathrm dx^2} - 2x\frac{\mathrm dP}{\mathrm dx} + \left[\ell\left(\ell+1\right)- \frac{m^2}{1-x^2}\right]P = 0 \tag{02} $$

Awesome! Now how on earth do I solve this? with the extra $\frac{m^2}{1-x^2}$ term?

I guess I could multiply by that on both sides and get rid of it from the denominator but then is it still Legendre equation? Have I read something wrong?

Thank you for your help and as always apologies if I missed something obvious and the question is silly!

ANSWER: Here's the link to a step by step solution for the general Legendre equation: http://www.physicspages.com/2011/03/22/associated-legendre-functions/

Answer 1

After searching for a while I finally found a detailed solution to this general Legendre equation.
So, I guess this question is answered!

The source for the solution is this:
http://www.physicspages.com/2011/03/22/associated-legendre-functions/

EDIT: As of February 12, 2018 the above link doesn't work. Use this instead: https://web.archive.org/web/20170509204654/http://www.physicspages.com:80/2011/03/22/associated-legendre-functions/

Answer 2

The page posted above by SilverSlash has been removed from the Wayback Machine but if you make the substitution $y=(1-x^2)^{m/2}f(x)$ you get another differential equation

$\Large \left(1-x^2\right) \frac{d^2f(x)}{dx^2} -2(m+1)x \frac{df(x)}{dx} + \left(A - m(m-1)\right)f(x) = 0$

which you can solve using a basic power series approach.

See here.