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

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.

Using the general solution in part (a), the initial condition givesTherefore the solution of this initial value problem isThe formula is used on the interval containing $x=1$ where the original ordinary differential equation and the displayed expression are defined.

The zero solution satisfies both the ordinary differential equation and the initial condition. The right-hand side is continuously differentiable in $y$ near $(x,y)=(2,0)$, so the Picard-Lindelöf theorem gives local uniqueness; equivalently the nonzero family in part (a) can never take the value zero at $x=2$. ThusThe restriction $x>0$ is the maximal interval containing $2$ on which the original ordinary differential equation is defined.