Use the product rule in a homogeneous-looking substitution

$dy/dx=v+x\,dv/dx$. Substitution gives $x(v+xv\prime)=vx+x^2$, so $x^2v\prime=x^2$. On an interval not containing zero, $dv/dx=1$ and $v=x+C$. This transformed equation is solved by direct integration, as highlighted by the teacher.

$y=xv=x^2+Cx$. At $x=-1$, $0=1-C$, hence $C=1$ and $y=x^2+x$. The substitution is valid on the interval $x<0$ containing the initial point. The resulting polynomial also directly satisfies the original equation at x=0 and extends to all real x; this extension is checked rather than inferred by dividing by zero.