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 / Solution

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.