= Solution
Write $Y(t)=y(e^t)$, so $x=e^t>0$. The <chain rule> gives
$$
Y'(t)=xy'(x),\qquad Y''(t)=xy'(x)+x^2y''(x).
$$
Hence $x^2y''=Y''-Y'$ and the <Cauchy-Euler equation> becomes
$$
\boxed{Y''+3Y'+2Y=0}.
$$
Its <characteristic equation> is $r^2+3r+2=(r+1)(r+2)=0$. The distinct roots give $Y(t)=Ae^{-t}+Be^{-2t}$. Returning to the original independent variable gives
$$
\boxed{y(x)=\frac A x+\frac B{x^2}\quad(x>0)}.
$$
The same form holds on any negative interval avoiding $0$, using $t=\log|x|$; the stated substitution covers the positive interval. At $x=0$ the <differential equation> is singular, so crossing it is a separate extension problem.
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