I took a test in which I was asked to give a solution in terms of a power series for the equation: $$x^2y''(x)-xy'(x)+(1-x)y(x)=0,~~~~~~~x>0.$$
At first I began to work on a power series centered at $0$ but then realised that the function is not defined at $0$. So I centered the solution at $x_0\in(0,\infty)$. My question is, even if I can give a power series around $0$ as a solution, is this solution valid? My guess is that it is not because $0$ is not even in the radius of convergence of the solution.
I appreciate your thoughts.
$0$ is a regular singular point of this differential equation. The indicial equation is $r^2 - 2 r + 1 = 0$, so there is a double indicial root $r=1$. This means there will indeed be a series solution of the form $$ y(x) = x + \sum_{j=2}^\infty a_j x^j $$ as well as one containing logarithmic terms $$ y(x) = x \log(x) + \sum_{j=2}^\infty (a_j \log(x) + b_j) x^j $$ The first is valid for all $x$ in some interval around $0$ (including $0$ itself), the second in an interval $0 < x < R$.