= Solution
Factor the right-hand side as $f(y)=y(y-2)(y+1)$. Its <equilibrium points> are $-1,0,2$, and $f'(y)=3y^2-2y-2$ gives
$$
\boxed{y=-1:\ \text{unstable};\qquad y=0:\ \text{asymptotically stable};\qquad y=2:\ \text{unstable}.}
$$
Indeed $f'(-1)=3$, $f'(0)=-2$ and $f'(2)=6$. The <phase line> has downward motion for $y<-1$, upward motion for $-1<y<0$, downward motion for $0<y<2$, and upward motion for $y>2$. The uniqueness of a smooth <autonomous differential equation> prevents trajectories from crossing an <equilibrium point>. Consequently trajectories starting between $-1$ and $2$ approach zero as $x\to\infty$, while those outside that interval move away from the three <equilibrium points>. The constant trajectories and representative nonconstant trajectories are shown below.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-2-scalar-solutions.png]
{title=Representative trajectories and equilibrium levels for the scalar cubic differential equation}
Back to article page
