Solution
= Solution
Write $Y(t)=y(e^t)$, so $t=\log x$. The <chain rule> gives
$$
y'(x)=\frac{Y'(t)}x,\qquad
y''(x)=\frac{Y''(t)-Y'(t)}{x^2}.
$$
Substitution into the <Cauchy-Euler equation> therefore gives $Y''+Y'-2Y=0$. Its <characteristic equation> factors as $(r-1)(r+2)=0$, so $Y=C_1e^t+C_2e^{-2t}$. Returning to $x$ gives
$$
\boxed{y(x)=C_1x+C_2x^{-2},\qquad x>0.}
$$