Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 15 / a / ii / Solution

For $x\ne 0$ and $y\ne 0$, put $v=y/x$. This is the standard substitution for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, the equation becomesTaking antiderivatives yields $v^2=2\ln |x|+C$. The explicit general solution is thereforeon any interval where $x\ne 0$ and $2\ln |x|+C>0$, with a fixed choice of sign. For a check that does not divide by $y$, set $w=y^2$ to get $w^{\prime }-2w/x=2x$ and hence $w=x^2(2\ln |x|+C)$.

There is no missing zero solution: at $y=0$ with $x\ne 0$, the original equation would require $x^2=0$. Also no real solution on an interval about $x=0$ can arise from this family, since $2\ln |x|+C$ becomes negative near $0$.