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/

  • 2
    $\begingroup$ 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 $\endgroup$
    – zeldredge
    Sep 6, 2015 at 14:54

2 Answers 2


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:

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/

  • $\begingroup$ You should probably post that specific solution as an answer; otherwise this post will probably be deleted as Not An Answer. Adding the solution would also help future readers of your question. $\endgroup$
    – HDE 226868
    Sep 6, 2015 at 17:58

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.


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