 Solution

Write $Y(t)=y(e^t)$, so $x=e^t>0$. The chain rule givesHence $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$ and the Cauchy-Euler equation becomesIts characteristic equation is $r^2+3r+2=(r+1)(r+2)=0$. The distinct roots give $Y(t)=A{e}^{-t}+B{e}^{-2t}$. Returning to the original independent variable givesThe 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.