Frobenius series for 2xy'' + xy' - (x+1)y = 0

[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?