Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 13 / a

Let $M=2x+e^y$ and $N=x{e}^y-\cos y$. Their cross partial derivatives agree: $M_y=e^y=N_x$. On the simply connected plane, this makes the differential form exact. Integrating $F_x=M$ gives $F=x^2+x{e}^y+h(y)$; matching $F_y=N$ then requires $h^{\prime }(y)=-\cos y$, so one potential isThe ordinary differential equation states $dF/dx=0$ along a solution. Its initial condition fixes the constant to $F(1,\pi /2)=e^{\pi /2}$. Hence the solution is specified implicitly byAt the initial point $F_y=e^{\pi /2}\ne 0$, so the implicit function theorem guarantees a unique local branch $y(x)$ through that point.