I have the ODE: $\displaystyle y''(x)+\frac{y'(x)}{x+1}+y(x)=0$
I know that this has a regular singular point at $x=-1$, as $(1+x)^{-1}$ has only a first order pole, and $1$ has no pole at all, and therefore I expect that it should have a non-singular (finite) solution (or so Riley,Hobson,Bence tells me).
Yet, the solution:
$\displaystyle y(x)=\frac{Y_0(2) J_0(x+1)-J_0(2) Y_0(x+1)}{J_0(1) Y_0(2)-J_0(2) Y_0(1)}$
tends to $\infty$ as $\rightarrow -1$. I'm therefore confused... can anybody resolve why this is?