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

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.