Probing series solution of DE: irregular singular point

[K29]

Homework Statement

Consider
$y''+\frac{\alpha}{x^{r}}y'+\frac{\beta}{x^{s}}y=0$
Suppose we try find a solution of the form<br />
$y=\sum_{n=0}^{\infty}a_{n}x^{n+\lambda}$
Show that if r=2 and s=1 then there is only one possible value of $\lambda$ for which there is a formal solution in the form shown above.

The Attempt at a Solution

Substituting the general solution into the DE gives us:

$\sum_{n=0}^{\infty}(n+ \lambda )(n+ \lambda -1)a_{n}x^{n+ \lambda -2}+\frac{\alpha}{x^{2}}\sum_{n=0}^{\infty}(n+ \lambda )a_{n}x^{n+ \lambda -1}+\frac{\beta}{x}\sum_{n=0}^{\infty}a_{n}x^{n+ \lambda }=0$

$\sum_{n=0}^{\infty}(n+\lambda)(n+\lambda-1)a_{n}x^{n+ \lambda -2}+\alpha \sum_{n=0}^{\infty}(n+ \lambda )a_{n}x^{n+ \lambda -3}+\beta \sum_{n=0}^{\infty}a_{n}x^{n+ \lambda -1}=0$

Pulling out some terms to make the starting terms in the summation have the same order of x:
$\lambda(\lambda-1)a_{0}x^{\lambda-2}+\sum_{n=1}^{\infty}(n+\lambda)(n+\lambda-1)a_{n}x^{n+\lambda-2}+\alpha \lambda a_{0}x^{\lambda-3}+\alpha(1+\lambda)a_{1}x^{\lambda-2 }+\alpha \sum_{n=2}^{\infty}(n+\lambda)a_{n}x^{n+\lambda-3}+\beta \sum_{n=0}^{\infty}a_{n}x^{n+\lambda-1}=0$


Short of shifting the summations and doing some grouping(I don't feel like that is necessary to complete the answer to the question), at this point I'm not sure what I should do. Do I look at the coefficients of the lowest power of x as I would do if there was a quadratic indical equation? If that were the case I would simply say$\alpha \lambda a_{0}=0$ therefore the only possible value for $\lambda$ is zero(since alpha and $a_{0}$ are arbitrary). But then what does that mean for all the other co-efficients? Why would I be interested in that term anyway, since it is not actually an indical equation? I'm pretty sure that the solution lies in TRYING to get a quadratic indical equation and consequently failing

I feel like I'm not far from deducing the answer, but I'm uneasy since I think I'm essentially dealing with a question that is telling me to 'show that the Frobenius method won't work for an irregular singular point'. I have no rules for trying to do this.

Note: There is a similar question in Boyce & DiPrima Elementary Differential Equations and Boundary Value Problems(2nd edition)

 

[lippyka]

Chapter 8 problem 1a, but unfortunately the solution is not given.Any help would be greatly appreciated, thank you.