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I have already shown that the equation has a regular singular point at $x=0$ and started using the Frobenius method which is the method that I am supposed to use to answer this question. So far I have got it to the point

\begin{multline} \sum_{n=1}^{\infty}(n+\alpha+1)(n+\alpha+2)a_{n-1}x^{n+\alpha-1}-\sum_{n=0}^{\infty}(n+\alpha)(n+\alpha-1)a_nx^{n+\alpha-1}+\\\frac72\sum_{n=1}^{\infty}(n+\alpha-1)a_{n-1}x^{n+\alpha-1}-3/2\sum_{n=0}^{\infty}(n+\alpha)a_nx^{n+\alpha-1}+\frac32\sum_{n=1}^{\infty}a_{n-1}x^{n+\alpha-1}, \end{multline}

But I'm not sure how to get the first terms out and combine the rest of the terms.

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I think the first term might be $-alpha(alpha-5/2)a_0x^{alpha-1}$ but i'm not completely sure. – Adam Nov 11 '12 at 16:15
up vote 0 down vote accepted

You need to drop the zero terms from the second and fourth sums, and then you will have something like $$ \big({}\big) a_ 0 x^{\alpha-1} + \sum_{n=1}^\infty \big({}\big)x^{n+\alpha-1} $$ The term multiplying $a_0 x^{\alpha-1}$ is the indicial equation.

Can you take it from here?

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I got $-3/2{\alpha}a_0x^{\alpha-1}$ and -{\alpha}({\alpha}-1)a_0x^{\alpha-1} but am unsure how to combine them to get a single term for a_0 – Adam Nov 11 '12 at 17:22
@Adam Why is this a problem? You have $$-\left(\alpha(\alpha - 1) +\frac{3}{2}\alpha\right)a_0 x^{\alpha-1} + \sum_{n=1}^\infty \big({}\big)x^{n+\alpha-1} = 0.$$ Given the orthogonality of the set $\{x^k\}_{k=0}^\infty$, the only way for the equation to be satisfied for all $x$ is that all constant terms are zero, in particular the one outside the sum, i.e. $$\alpha(\alpha + \tfrac{1}{2}) = 0.$$ This is the indicial equation. Note that $a_0 = 0$ is also a posibility but, you can assume without loss of generality that this doesn't happen (why?). – Pragabhava Nov 11 '12 at 17:31
because if that was the case you would just get y=0 which is trivial – Adam Nov 11 '12 at 17:41
@Adam In this particular case what you've said is true, but that doesn't happen always. Let $y = \sum_0^\infty a_n x^{n + \alpha}$. If $a_0 = 0$, the lowest power of the ansat would be $\alpha + 1$. Redefining $\beta = \alpha + 1$ and $b_n = a_{n+1}$, $y = \sum_0^\infty b_n x^\beta$ and you have the same ansat as before. Hence, one can assume without loss of generality that $a_0 \neq 0$. – Pragabhava Nov 11 '12 at 17:54
aah that makes sense, thanks for the help – Adam Nov 11 '12 at 18:06

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