Past exam of the mathematical methods I course of the University of Oxford / 2019 / 3 / Solution

The right-hand side depends on $x+y$, suggesting the substitution $u=x+y$. Then $y^{\prime }=u^{\prime }-1$, so the ordinary differential equation becomesThe original equation is undefined at $u=0$. If $u\ne 2$, this is a separable differential equation:Substituting back gives the implicit familyDifferentiating the implicit equation gives $y^{\prime }-1+2(1+y^{\prime })/(x+y-2)=0$, or $(x+y)y^{\prime }=x+y-4$, verifying it on its permitted intervals.

Division by $u-2$ excluded the constant solution $u=2$, which must be restored separately:It satisfies $y^{\prime }=-1=(2-4)/2$. Together these give all local real solutions on intervals where $x+y\ne 0$.