Using Frobenius, show $y_1 = x$ of eqn $x^{3}y''-xy'+y=0$

Question

My question is right at the beginning, I would like to use the indicial equation but solving for limits $p_0$ and $q_0$ for $r(r-1)+p_0r + q_0 = 0$, ie:

$p_0=\lim xp(x)$ for $x\rightarrow 0$ is $\infty$

What am I doing wrong?

Answer 1

We may start by noticing that $y(x)=x$ is a solution of the given differential equation, and any solution of the given differential equation has to fulfill $y(0)=0$. By setting $y(x)=x\cdot f(x)$ the given DE boils down to:

$$ x^3 (2f'(x)+ x f''(x) ) - x (f(x)+ x f'(x)) + x f(x) = 0 $$ or: $$ x^3 f''(x) + (2x^2- x)\,f'(x) = 0 $$ that is a separable DE for $f'$.
It follows that the solutions to the original DE are given by $y(0)=0$ and $$ y(x) = A x + B x e^{-1/x}\quad \text{for }x\neq 0. $$ If $B\neq 0$, there is an essential singularity at the origin, hence the Frobenius method is not well-suited for solving this problem. And actually, if $B\neq 0$, $y(x)$ is unbounded in a left neighbourhood of the origin.