User:Arthur Rubin/Reference desk

Reference desk/Archives/Mathematics/2011 November 28
Let $$2x(x-1)y'' + 3(x-1)y' - y = 0$$. This has a regular singular point at $$x = 0$$.


 * I get, for the $$x^{n-1+r}$$ term,
 * $$\left( 2(n+r-1)(n+r-2) + 3 (n+r-1) - 1\right) a_{n-1}(r) = \left( 2 (n+r-1)(n+r) + 3 (n+r) \right)a_{n}(r)$$, so the indicial equation is the rhs coefficient is 0 at $$n=0$$, which reduces to $$r(2r+1)=0$$. The recursion is:
 * $$a_n(r) = \frac {n+r - \tfrac 3 2}{n+r+\tfrac 1 2} a_{n-1}(r)$$

The solution for $$r=-\tfrac 1 2$$ is
 * $$y = x^{-\tfrac 1 2}(1-x)$$

A solution for $$r=0$$ is

$$\begin{align} y & = \sum_{n=0}^\infty \frac 1 {(n+\tfrac 1 2)(n-\tfrac 1 2)}x^n \\ & = \sum_{n=0}^\infty \left(\frac 1 {n-\tfrac 1 2} - \frac 1 {n + \tfrac 1 2} \right) x^n \\ & = -\frac {1} {2} - (1-x) \sum_{n=0}^\infty \frac 1 {n + \tfrac 1 2} x^n \\ & = -\frac 1 2 - (1-x) x^{-\tfrac 1 2}\int _0 ^x \frac {t^{- \tfrac 1 2}}{1-t}\, dt \text { only correct for }x > 0,\text{ at best}\\ & = -\frac 1 2 - (1-x) x^{-\tfrac 1 2} \ln \frac {1+x^{-\tfrac 1 2}} {1-x^{-\tfrac 1 2}} \\ \end{align}$$

This is probably not correct