Past exam of the mathematical methods I course of the University of Oxford / 2013 / 6 / Solution

Here the dependent variable is $x$ and the independent variable is $y$. Multiplying the equation by $2x$ and setting $q=x^2$ gives a linear equation:An integrating factor is $y^{-4}$, soThus $x^2=C{y}^4-y^2$. The initial point $(x,y)=(1,1)$ fixes $C=2$, and the positive branch through that point isEquivalently, inverting this branch givesThe source equation contains $xy$ in the denominator. The endpoint with $x=0$ is therefore excluded even though the inverse expression has a finite limit there. A negative-$x$ branch cannot be joined through that endpoint to produce a solution $x(y)$ of the original equation. The interval shown is the maximal real positive branch containing the initial point.