 Solution

For $x\ne 0$, divide by $x^2$ to obtain the linear ordinary differential equation $y^{\prime }+3y/x=x^{-2}$. An integrating factor is $x^3$ on either interval not containing zero. Thus $(x^{3}y{)}^{\prime }=x$ andSubstituting gives $x^2{y}^{\prime }+3xy=1$ directly. No classical solution can include $x=0$, since the original equation there would assert $0=1$; constants on the two separate intervals need not agree.