Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 16 / a / ii / Solution

Here $P=y^2$ and $Q={\cos }^{2}x+y\sin 2x$. Their derivatives giveThus on intervals with $\cos x\ne 0$, choose the integrating factor $\mu =e^{\int 2\tan xdx}={\sec }^{2}x$. The multiplied form isConsequentlyThis implicit family includes $y=0$ and the branch $y=-\cot x$ when $C=0$. Where the coefficient of $y^{\prime }$ is nonzero, its differentiated form directly recovers the original equation. A useful equivalent relation is $y\cos x+y^2\sin x=C\cos x$. At $\cos x=0$, the original equation requires $y=0$; continuation there must be checked in that original equation rather than justified by the singular integrating factor. In particular the zero solution extends everywhere, and $y=-\cot x$ extends smoothly across the cosine zeros where it is defined.