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

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.

Let $P=\mu (x)(xy-16x-x^3)$ and $Q=\mu (x)(16+x^2)$. An exact differential form must satisfy $P_y=Q_x$. These partial derivatives giveTaking antiderivatives produces $\ln |\mu |=-\frac{1}{2}\ln (16+x^2)+C$. A convenient nonzero integrating factor for a differential one-form is thereforeAn explicit potential, which also proves global exactness, isIndeed $F_x=P$ and $F_y=Q$ for the chosen integrating factor.

Multiplying the ordinary differential equation by the nonzero integrating factor from part (i) makes its left-hand side $dF(x,y(x))/dx$. Hence the potential is constant:Since $16+x^2>0$ for every real $x$, this is equivalent to the explicit general solutionIt is defined for all real $x$. There are no lost solutions, since the integrating factor and the coefficient of $y^{\prime }$ are everywhere nonzero.