Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 13 / a / Solution

On an interval with $x\ne 0$, the equation is a separable differential equation. For $y\ne 0$,Taking antiderivatives gives $-1/y=-4/x-x^3/3+C$. Renaming the constant, the explicit general nonzero solution isIt is defined on any interval avoiding $x=0$ and zeros of its denominator. The separation divided by $y^2$, so the additional solution $\boxed{y(x)\equiv 0}$ must be included.