Solution
= Solution
For $x\ne0$, divide by $x^2$ to obtain $y'+y/x=1/x^2$, a first-order <linear ordinary differential equation>. An <integrating factor> is $x$ on either interval not crossing zero, since the <product rule> gives
$$
(xy)'=\frac1x.
$$
Integration and division by $x$ give the explicit <general solution>
$$
\boxed{y(x)=\frac{\ln|x|+C}{x},\qquad x\ne0.}
$$
The original equation would read $0=1$ at zero, so no solution can include that point.