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