Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 13 / a

Work on an interval with $x\ne 0$. The homogeneous first-order differential equation becomes a separable differential equation under $v=y/x$: substituting $y=xv$ gives $v+x{v}^{\prime }=v^2+v$, hence $x{v}^{\prime }=v^2$.

For $v\ne 0$, integration gives $-1/v=\ln |x|+C$. Renaming the arbitrary constant producesFor the nonzero family the denominator must remain nonzero, so take any connected interval avoiding both $x=0$ and $|x|=e^C$. The zero solution must be restored after division by $v^2$; by the Picard-Lindelöf theorem away from $x=0$, a solution cannot switch between zero and a nonzero branch there.