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

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/

• Your first equation is valid in cases of spherical symmetry. You'll need to use associated Legendre polynomials for the more general equation: en.wikipedia.org/wiki/Legendre_function – zeldredge Sep 6 '15 at 14:54