Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 18

If the form is $dF$ for a scalar function with continuous second partial derivatives, then $p=F_x$ and $q=F_y$. Equality of the mixed partial derivatives therefore requiresThis is necessary for an exact differential. For continuously differentiable coefficients on a simply connected open domain it is also sufficient; on a domain with holes, the equality alone need not give a single-valued global potential.

For the printed coefficients, $p=2\sin x\cosh y$ and $q={\sin }^{2}x\sinh y$. Their cross derivatives areThey are not equal as functions; their difference is $2\sin x(1-\cos x)\sinh y$. Hence the form is not exact. Agreement on certain lines or isolated points is insufficient, because exactness requires the derivative identity throughout an open domain.

Let $w=x+iy$. Then $y-ix=-iw$, and the complex trigonometric identity $\sin (-iw)=-i\sinh w$ givesThus the form is exact, with complex-valued scalar potential $F(x,y)=\cosh (x+iy)$. Differentiating that potential with respect to the two real variables checks both coefficients individually.

Away from zeros of $\sin x$, an integrating factor for a differential one-form is $\mu (x)=1/\sin x$. Multiplication converts the form toTherefore the implicit solution family is $xy+\ln |\sin x|=C$. On each graph interval between consecutive multiples of $\pi$, this becomesNo such open interval contains zero. At $x=k\pi$, the undivided graph equation would read $\cos (k\pi )=0$, which is impossible for a finite derivative; hence no differentiable graph extends across one of these points.

If solutions are instead understood as arbitrary parametrized integral curves of the differential form, the vertical lines $x=k\pi$ are additional solutions: along them $dx=0$ and $x\sin xdy=0$. They were lost on division by $\sin x$ and are not graphs $y(x)$ over an interval.

View $a=a(b,c)$ with $b=b(d,e)$ and $c=c(d,e)$. Hold $e$ fixed and apply the multivariable chain rule:Dividing the changes along the fixed-$e$ variation by $dd$ gives exactly the two terms in the question. Thus the simpler expression isThe subscripts specify which coordinate is held fixed and must be retained.