Consider the differential equation below. (You do not need to solve this differential equation to answer this question.) y' = y^2(y + 4)^3 Find the steady states and classify each as stable, semi-stable, or unstable. Draw a plot showing some typical solutions. If y(0) = -2 what happens to the solution as time goes to infinity?

We have [tex]y'=0[/tex] when [tex]y=0[/tex] or [tex]y=-4[/tex], so we need to check the sign of [tex]y'[/tex] on 3 intervals:Suppose [tex]-\infty<y<-4[/tex]. In particular, let [tex]y=-5[/tex]. Then [tex]y'=(-5)^2(-5+4)^3=-25<0[/tex]. Since [tex]y'[/tex] is negative on this interval, we have [tex]y(t)\to-\infty[/tex] as [tex]t\to\infty[/tex].Suppose [tex]-4<y<0[/tex], say [tex]y=-1[/tex]. Then [tex]y'=(-1)^2(-1+4)^3=-27<0[/tex], so that [tex]y(t)\to-4[/tex] as [tex]t\to\infty[/tex].Suppose [tex]0<y<\infty[/tex], say [tex]y=1[/tex]. Then [tex]y'=1^2(1+4)^3=125>0[/tex], so that [tex]y(t)\to\infty[/tex] as [tex]t\to\infty[/tex].We can summarize this behavior as in the attached plot. The arrows on the [tex]y[/tex]-axis indicate the direction of the solution as [tex]t\to\infty[/tex]. We then classify the solutions as follows.[tex]y=0[/tex] is an unstable solution because on either side of [tex]y=0[/tex], [tex]y(t)[/tex] does not converge to the same value from both sides.[tex]y=-4[/tex] is a semi-stable solution because for [tex]y>-4[/tex], solutions tend toward the line [tex]y=-4[/tex], while for [tex]y<-4[/tex] solutions diverge to negative infinity.