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

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.