Solve Frobenius Series Homework for 2 Independent Solutions

[Ted123]

Homework Statement

The origin is a regular singular point of the equation 2x^2 y'' + xy' - (x+1)y =0. Find 2 independent solutions which are Frobenius series in x.

The Attempt at a Solution

Substituting y = \sum_{n=0}^{\infty} a_n x^{n + \sigma} eventually gives (2\sigma(\sigma - 1) +\sigma -1 )a_0 x^{\sigma} + \sum_{n=0}^{\infty} \left[ (2(\sigma + n)(\sigma + n+1) + \sigma + n ) a_{n+1} - a_n \right] x^{n+\sigma + 1} = 0.

Equating the series to 0 term-by-term gives the indicial equation 2\sigma (\sigma -1) + \sigma -1 = 0 \Rightarrow (2\sigma +1)(\sigma -1) = 0 \Rightarrow \sigma = -\frac{1}{2},\; \sigma = 1.

We get the recurrence relation a_{n+1} = \frac{a_n}{2(\sigma + n)(\sigma + n +1) + \sigma + n}.

This is what I'm struggling to solve...

 

[Ray Vickson]

Don't you have the solution now? You know there are two possible values of sigma. For each sigma you have a slightly different recursion, leading to a slightly different series. In general, you have \displaystyle a_1 = \frac{a_0}{2 \sigma (\sigma +1) + \sigma} \:,
\displaystyle a_2 = a_0 \prod_{i=0}^{1} \frac{1}{2(\sigma+i)(\sigma+i+1) + \sigma+i},
etc.

RGV

 

[Ted123]

So the 2 independent solutions, for \sigma =1, -\frac{1}{2} is:

\displaystyle a_n = a_0 \prod_{i=0}^{n-1} \frac{1}{2(\sigma+i)(\sigma+i+1) + \sigma+i}\;?

Is there any significance to the question mentioning that x=0 is a regular singular point?