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Here is the statement of the problem:

Consider the Legendre Equation

$$ (*)\qquad (1-x^2)y''-2xy'+2y=0 $$

(a) Find two linearly independent solutions about $x=0$, solving completely any relevant recurrence relations.

(b) Compute the radius of convergence for each fundamental solution in part (a).

(c) Show that $x=1$ is a regular singular point of $(*)$.

(d) Assume that $y=(x-1)^s\sum_{n=0}^\infty a_n(x-1)^n$ (with $s \in \mathbb R$) is a solution near $x=1$, solve the indicial equation satisfied by $s$. Find one fundamental solution of $(*)$ near $x=1$.

Where I am:

Ok, this problem has been bugging me for a few days now. First off...

For (a): Since $x=0$ is an ordinary point of $(*)$, we may assume that a general solution of our ODE about this point is of the form:

$$ y=\sum_{n=0}^\infty a_nx^n. $$

So, taking some derivatives, plugging into our ODE, I found the following recurrence relation (assuming all of my arithmetic was correct):

$$ a_n = \frac{a_{n-2}(n^2-7n+12)}{n(n-1)} $$

Now, note that $a_0$ and $a_1$ are undefined here due to a zero in the denominator. Ok, fine. So, I started with $a_2$:

$$ a_2 = \frac{a_{0}(4-14+12)}{2} = -a_0 $$

Ok, sure. It's easy to see that, beyond that, all other terms vanish. Therefore, our solution becomes the following:

$$ y=\sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 = a_0 + a_1x -a_0x^2 = a_0(1-x^2)+a_1x $$

Sure, that seems fine; it's a nice little polynomial. But, how do I find the other solution? I'm completely clueless here.

Now, for part (b). Well, the radius of convergence for the solution that I found is infinity, I suppose, because it's just a polynomial. I suppose that once I find the other solution, this part will make more sense.

For part (c): This is easy. Just take some limits, and whatnot. I only included it here for context.

I haven't started part (d) yet because I'm too frustrated with the first part. I'm going to work on it now, though.

If anyone can I help me out here, I'd greatly appreciate it. Thanks!

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    $\begingroup$ For part (a), you have no condition on either $a_0$ or $a_1$. They are therefore independent of each other. So your expression is actually two solutions in one already. Congratulations! $\endgroup$ – bob.sacamento Jul 1 '15 at 18:14
  • $\begingroup$ Oh... you're right! Look at that. Hmm. The problem is that my instructor said that one function would be a polynomial and that the other would not be analytic, so I must've done something wrong... $\endgroup$ – thisisourconcerndude Jul 1 '15 at 18:19
  • $\begingroup$ Yep. I think $1-x^2$ is not a solution. $\endgroup$ – bob.sacamento Jul 1 '15 at 21:32
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$$(1-x^2)y''-2xy'+2y=0$$ Since $y=x$ is an obvious solution, we look for other solutions on the form $y=xf(x)$

$y'=f+xf'$ and $y''=2f'+xf''$

$(1-x^2)(2f'+xf'')-2x(f+xf')+2xf=0$ $$2(1-2x^2)f'+x(1-x^2)f''=0$$

$\frac{f''}{f'}=-2\frac{1-2x^2}{x(1-x^2)}$

The integration leads to : $$f'=c_1\frac{1}{x^2(1-x^2)}$$ Integrating again leads to : $$f=c_1\left(-\frac{1}{x}+\frac{1}{2}\ln\left| \frac{1-x}{1+x} \right| \right)+c_2$$ $$y=c_1 \left(-1+\frac{x}{2}\ln\left| \frac{1-x}{1+x} \right| \right)+c_2 x$$

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