Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 9 / c

Rearranging gives $y^{\prime }=2x(y-1{)}^2$. First retain the equilibrium $y=1$. For every other solution, separation of variables gives $\int (y-1{)}^{-2}dy=\int 2xdx$, so $-1/(y-1)=x^2+C$. HenceA nonconstant solution is defined on any interval where its denominator is nonzero. Dividing by $(y-1{)}^2$ without retaining the equilibrium would lose one solution.