Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 7 / a / Solution

With $M=2x{y}^2+4$ and $N=2x^{2}y-6$, the cross derivatives agree: $M_y=N_x=4xy$. This is an exact first-order ordinary differential equation. Integrating $M$ with respect to $x$ gives a potential $x^2{y}^2+4x+g(y)$; matching $N$ gives $g^{\prime }=-6$. The implicit solution is