Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 14 / b

With $v=x+y$, the chain rule givesFor $v\ne 2$, this separable differential equation givesRestoring $v$ and absorbing constants, the nonconstant branches satisfyThe constant branch lost in division is $v=2$, giving the additional solution $\boxed{y=2-x}$. All branches must avoid $v=4/3$, where the original differential equation is undefined.

Yes: $y=2-x$. Its derivative is $-1$, while substitution into the right-hand side gives $-2/(6-4)=-1$. This is the equilibrium $v=2$ of the transformed autonomous differential equation; it must be restored separately after dividing by $v-2$ in the separated solution.

It is the only straight-line solution. Substituting $y=mx+d$ into the original equation shows that $m\ne -1$ would require $m=-1/3$ from the $x$ coefficients, but the constant coefficients would then give $4/3=0$. For $m=-1$, the constant equation gives $d=2$.