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

For a continuously differentiable differential one-form, exactness requires $\boxed{P_y=Q_x}$. It is locally sufficient, and globally sufficient on a simply connected domain; on a domain with holes, closedness alone need not imply exactness.

An integrating factor for a differential one-form depending only on $x$ must satisfyThus, where $Q\ne 0$,If the ratio depends only on $x$, this defines a nonzero factor and makes it a closed differential form, hence locally exact. For a factor depending only on $y$, the analogous calculation givesprovided the ratio is a function only of $y$ on the region where $P\ne 0$. A nonzero constant multiple of either factor works equally well.

Write the equation as $Pdx+Qdy=0$, with $P=3x+y$ and $Q=2x$. Thenso a suitable integrating factor is $\mu =|x{|}^{-1/2}$ on either half-line. For example, on $x>0$ the resulting exact form isHence the solution, written to cover either sign of $x$, isThe original equation remains meaningful at $x=0$, where it requires $y=0$. Only $C=0$ gives a finite differentiable extension across zero, namely $y=-x$.

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.