Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 14 / c

Here $M_y-N_x=(6xy+2)-(4xy+1)=2xy+1$, whereas $N=x(2xy+1)$. Thus the ratio in (b) is $1/x$, and one may choose the nonzero integrating factor $\boxed{\mu (x)=x}$ on either half-line. A negative constant multiple on a half-line is harmless.

The multiplied form isHence $x^3{y}^2+x^{2}y=C$, a quadratic equation for $y$. Solving it givesChoose a real branch with positive discriminant to obtain a differentiable graph; a zero discriminant produces a vertical tangent. Although the ratio derivation initially cancels $2xy+1$, the displayed exact differential verifies the multiplying factor directly wherever $x\ne 0$. At $C=0$ the branches are $y=0$ and $y=-1/x$. Only $y=0$ has a finite smooth extension across $x=0$, where the original equation requires $y=0$.