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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.2012-11-11

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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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    aah that makes sense, thanks for the help2012-11-11