Solving associated Legendre equation

[ShayanJ]

I'm trying to solve the associated Legendre differential equation:

<br /> y&#039;&#039;+\frac{2x}{x^2-1}y&#039;+[ \lambda+\frac{m^2}{x^2-1}]y=0<br />

By series expansion around one of its regular singularities.(e.g. x_0=1)

This equation is of the form:

<br /> y&#039;&#039;+p(x)y&#039;+q(x)y=0<br />

Which is solved by the following procedure:
1-Expand p(x) and q(x) in laurent series:

<br /> p(x)=\sum_{-1}^{\infty} A_n (x-x_0)^n \\ \\<br /> q(x)=\sum_{-2}^{\infty} B_n(x-x_0)^n<br />

2-Solve the equation r^2+(A_{-1}-1)r+B_{-2}=0 for r
3-Based on the value(s) of r, write y as a series and find the coefficients by substitution in the differential equation.

For the associated Legendre eqaution,I find A_{-1}=1 and B_{-2}=0
which means r=0.So the solutions have the forms:

<br /> y_1=1+\sum_1^{\infty}a_n(x-1)^n \\<br /> y_2=y_1 \ln{|x-1|}+\sum_0^{\infty}b_n(x-1)^n<br />

When I substitute y_1 in the equation,I get the following:

<br /> \sum_1^{\infty} [ n(n+1)a_n + \large{\frac{2xna_n}{x+1} +\frac{m^2 a_{n-1}}{x+1} }](x-1)^n+[\lambda(x-1)+\frac{m^2}{x+1}](x-1)=0<br />

Now I have two questions:
1-Its obvious that the relation between a_n and a_{n-1} involves x.Is it acceptable?why?
2-How am I supposed to use the equation,which is gained by equating the terms outside the sum to zero?

Thanks

 

[ShayanJ]

Well,I guess I should change my question a bit.
First,the differential equation I wrote in the last post isn't the associated legendre equation.
Second,I now understand its completely wrong to have x in the recursion relation.
So I should ask,what's wrong with my way of solving that equation?
Thanks