Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 16 / b / Solution

Construct the scalar potential $F(x,y)=y{e}^{x^2}-x^2$. Its differential iswhich proves exactness directly. Multiplying the differential equation by $e^{x^2}$ shows that a solution satisfies $dF/dx=0$, so $y{e}^{x^2}-x^2=C$. The initial value at zero gives $C=-1$. ThereforeThe factor $e^{x^2}$ is also an integrating factor for the associated first-order linear equation.