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

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 $y\equiv 0$.

For $x\ne 0$ and $y\ne 0$, put $v=y/x$. This is the standard substitution for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, the equation becomesTaking antiderivatives yields $v^2=2\ln |x|+C$. The explicit general solution is thereforeon any interval where $x\ne 0$ and $2\ln |x|+C>0$, with a fixed choice of sign. For a check that does not divide by $y$, set $w=y^2$ to get $w^{\prime }-2w/x=2x$ and hence $w=x^2(2\ln |x|+C)$.

There is no missing zero solution: at $y=0$ with $x\ne 0$, the original equation would require $x^2=0$. Also no real solution on an interval about $x=0$ can arise from this family, since $2\ln |x|+C$ becomes negative near $0$.