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 / i

Where $\cos x\ne 0$, the separable differential equation is $y^{\prime }=y\tan x$. Separating and taking antiderivatives gives $\ln |y|=-\ln |\cos x|+C$, for nonzero $y$. Including the zero solution, the explicit general solution ison any interval with $\cos x\ne 0$. It can also be obtained without dividing by $y$: the product rule gives $(y\cos x{)}^{\prime }=0$.

A nonzero solution cannot extend through a zero of $\cos x$. If a differentiable solution is required on an interval containing such a zero, the original equation there requires $y=0$, and the only such solution across that zero is $y\equiv 0$.