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I am interested in finding two linearly independent solutions of the equation

$$(1-x^2)y'' - 2xy' + \lambda y=0 \quad\quad (1)$$

using the Frobenius method (all the way through and not randomly deciding to take s=0 since that's just a regular power series [See http://mathworld.wolfram.com/LegendreDifferentialEquation.html for reference]).

We begin by writing out the needed derivatives

$$\begin{align}y =& \sum_{n=0}^{\infty} C_n x^{s+n} \\ y' =& \sum_{n=0}^{\infty} C_n (s+n) x^{s+n-1} \\ y' =& \sum_{n=0}^{\infty} C_n (s+n) (s+n-1) x^{s+n-2}\end{align}$$

and we substitute back into (1) as

$$(1-x^2)\sum_{n=0}^{\infty} C_n (s+n) (s+n-1) x^{s+n-2} - 2x\sum_{n=0}^{\infty} C_n (s+n) x^{s+n-1} + \lambda \sum_{n=0}^{\infty} C_n x^{s+n}=0. $$

If we expand the first term and simplify the second term, we can rearrange to get to

$$\sum_{n=0}^{\infty}\left[\lambda C_n - 2 C_n (s+n) - C_n(s+n)(s+n-1)\right]x^{s+n} + \sum_{n=0}^{\infty} C_n (s+n) (s+n-1) x^{s+n-2}=0$$

where we are left with an $x^{s+n-2}$ term. Typically in these problems, we would rewrite that term so that it can be combined with the $x^{s+n}$ terms, usually giving us a $C_{n+2}$ term that provides the needed recurrence equation. In a regular power series without the added power of $s$ this would simplify down nicely since we would get

$$ \sum_{n=0}^{\infty} C_n (n) (n-1) x^{n-2} \\ \text{Let } m = n-2 \Rightarrow n = m+2 \\ \sum_{m=-2}^{\infty} C_{m+2} (m+2) (m+1) x^{m}\\ \sum_{n=0}^{\infty} C_{n+2} (n+2) (n+1) x^{n}$$

where we can neglect the first two terms since $n=-2$ and $n=-1$ result in the term being zero. However, in this case, things do not simplify down nicely and we get

$$\sum_{n=-2}^{\infty} C_{n+2} (s+n+2)(s+n+1) x^{s+n}$$

where we aren't able to simplify down (that I can see). The next step after simplifying everything down would be to group together all of the terms and require that the product of the $C_1$ term and its coefficients be equal to zero. This will provide two values for $s$, which happen to be 0 and 1, effectively bringing us back to the original power series solution.

This problem is the same as Solution of the Legendre's ODE using Frobenius Method except that question explicitly writes out the sums and then states the indicial equation. I guess I'm missing the link between the two. How do I simplify this problem down to get to the indicial equation? I'm certain I'm missing something simple, but it's not quite clicking at the moment.

Thanks!

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I was definitely way overthinking this. I believe the solution is to instead shift the first term forward rather than shifting the second term back. This gets you to

$$\sum_{n=2}^{\infty}C_{n-2}\left[\lambda+2(s+n-2)-(s+n-2)(s+n-3)\right]x^{s+n-2} + \sum_{n=0}^{\infty}C_n (s+n)(s+n-1)x^{x+n-2} = 0$$

where we then obtain the coefficients for $C_0$ from the second term where we get

$$(s)(s-1) = 0 \Rightarrow s = 0, 1$$

Everything from here is as in Solution of the Legendre's ODE using Frobenius Method

I feel a bit silly now. I think staring at equations for too long is never a good idea.

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