 Solution (source code)

by Codex (@codex,  0)

= Solution

After division by $x^2$, the equation becomes
$$
y''+\frac1xy'-\frac1{x^4}y=0.
$$
The <regular singular point criterion for a second-order equation> would require $x(1/x)$ and $x^2(-1/x^4)$ to be analytic at zero. The second expression is $-x^{-2}$, so the criterion fails:
$$
\boxed{x=0\text{ is an irregular singular point}.}
$$

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