Past exam of the mathematical methods I course of the University of Oxford / 2013 / 12 / b

On an interval with $x\ne -1$, an integrating factor is $(x+1{)}^{-2}$. Multiplication givesIntegration therefore yieldsThe polynomial formula has a smooth continuation at $x=-1$, but the original coefficient $2/(x+1)$ is undefined there. Solution intervals for the stated equation must avoid that point, and the constants on its two sides need not agree.

The logarithm requires $x>0$ for real-valued solutions. Put $v=xy$. Since $v^{\prime }=x{y}^{\prime }+y$, the equation becomes $v^{\prime }=v^2\log x$. For a nonzero branch, separation givesRenaming the constant and restoring $y=v/x$ givesDividing by $v^2$ would omit the equilibrium $v=0$, so the full solution set also includes $\boxed{y\equiv 0}$. The nonzero branches are defined on connected intervals with $x>0$ and $C+x-x\log x\ne 0$. A regular solution cannot switch from a nonzero branch to zero at an ordinary point, because the right-hand side is locally Lipschitz in $v$ there. These domain restrictions are part of the full real solution.