= Solution
Under $y=xu$, the <product rule> gives $y'=u+xu'$ and $y''=2u'+xu''$. Substitute these into the original equation:
$$
x(2u'+xu'')-(u+xu')+u=x^2u''+xu'=x(xu''+u')=0.
$$
Since $x>0$, this is precisely $xu''+u'=0$, or $(xu')'=0$. Integrating twice gives $xu'=C_1$ and $u=C_1\ln x+C_2$. Therefore
$$
\boxed{y(x)=x(C_1\ln x+C_2),\qquad x>0.}
$$
The two independent <solutions of a differential equation> supply the <general solution> of this second-order <linear ordinary differential equation>.
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