Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 9 / c / Solution

When the two roots coincide, $y^{\prime }=-(y-a{)}^2$. Separation gives $y=a+1/(x+c)$, and the equilibrium $y=a$ must be added. Integrating $z^{\prime }=yz$ givesThe nonconstant pair is restricted to an interval excluding $x=-c$. Equivalently every nonzero linearizing function is $z=(A+Bx)e^{ax}$, with $y=a+B/(A+Bx)$ wherever its denominator is nonzero.