 Solution

Factor the right-hand side as $f(y)=y(y-2)(y+1)$. Its equilibrium points are $-1,0,2$, and $f^{\prime }(y)=3y^2-2y-2$ givesIndeed $f^{\prime }(-1)=3$, $f^{\prime }(0)=-2$ and $f^{\prime }(2)=6$. 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.