Question

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?

Answer 1

We have
`y'=0` when
`y=0` or
`y=-4`, so we need to check the sign of
`y'` on 3 intervals:

• Suppose
  `-\infty<y<-4`. In particular, let
  `y=-5`. Then
  `y'=(-5)^2(-5+4)^3=-25<0`. Since
  `y'` is negative on this interval, we have
  `y(t)\to-\infty` as
  `t\to\infty`.
• Suppose
  `-4<y<0`, say
  `y=-1`. Then
  `y'=(-1)^2(-1+4)^3=-27<0`, so that
  `y(t)\to-4` as
  `t\to\infty`.
• Suppose
  `0<y<\infty`, say
  `y=1`. Then
  `y'=1^2(1+4)^3=125>0`, so that
  `y(t)\to\infty` as
  `t\to\infty`.

We can summarize this behavior as in the attached plot. The arrows on the
`y`-axis indicate the direction of the solution as
`t\to\infty`. We then classify the solutions as follows.

• 
  `y=0` is an unstable solution because on either side of
  `y=0`,
  `y(t)` does not converge to the same value from both sides.
• 
  `y=-4` is a semi-stable solution because for
  `y>-4`, solutions tend toward the line
  `y=-4`, while for
  `y<-4` solutions diverge to negative infinity.