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

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

Here $M_y=2y$ whereas $N_x=0$, so the equation is not exact. The identity $M_y-N_x=N$ gives the integrating factor for a differential one-form $\mu (x)=e^x$. Multiplying makes the left-hand side the exact differential of $e^x(y^2-x+1)$. ThusFor real graph solutions, the right-hand side must be nonnegative; a smooth choice of square-root branch is made on the interval concerned. Finding the factor did not require dividing by $y$ at zero.

The cross derivatives are both $2y$, so the equation is exact. The function $x\cos x+x{y}^2$ differentiates to the given one-form. HenceDifferentiating this implicit equation checks both coefficients of the exact differential directly.