Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 1 / c

For this homogeneous first-order differential equation, set $v=y/x$, with $x\ne 0$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution simplifies the equation toThe useful further substitution is $w=e^{-v}-v^2$. Its chain rule derivative satisfiesThus $(x^{2}w{)}^{\prime }=0$. Replacing $v$ and $w$ gives the implicit solutionNo division by $w$ was needed, so the level $C=0$ is included. A direct verification also identifies the integrating factor for a differential one-form:Setting this exact differential to zero recovers precisely the original equation after multiplication by the nonzero factor $e^{y/x}$. The coefficient $1+2v{e}^v$ never vanishes for real $v$, since $v{e}^v\ge -1/e$ implies $1+2v{e}^v\ge 1-2/e>0$. Hence these level sets give local function graphs $y(x)$ everywhere with $x\ne 0$.