I have been solving Frobenius method problems. However it always has been a point of solution equal to 0 (x = 0). What happens when there's a given problem such as below and it's point of solution say x = 2?
$ x(x-1)y^{''}+(x-2)y^{'}-y = 0 $
Also some examples wherein point of solution (x = 0) is a singular point: LINK.
EDIT:
May I add a question, how do get now the singular points?