Question

use a computer system or graphing calculator to plot a slope field and/or enough solution curves to indicate whether each critical point of the given differential equation is stable, unstable, or semistable.
$\frac{dx}{dt}=(x^{2}-16)^{2}$
identify any stable critical point(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the differential equation has stable critical point(s) at x =
(simplify your answer. use a comma to separate answers as needed.)
b. the differential equation has no stable critical points

Explanation:

Step1: Find critical points

Set $\frac{dx}{dt}=(x^{2}-16)^{2}=0$. Then $x^{2}-16 = 0$, so $(x - 4)(x + 4)=0$. Solving gives $x=-4,4$.

Step2: Analyze stability using slope - field concept

For a critical point $x = c$, if solutions near $x = c$ move towards $x = c$ as $t$ increases, $x = c$ is stable; if they move away, it is unstable; if solutions move towards from one side and away from the other, it is semistable.
For $\frac{dx}{dt}=(x^{2}-16)^{2}=x^{4}-32x^{2}+256$. Consider the sign of $\frac{dx}{dt}$ in intervals:
Let's take test points. In the interval $x<-4$, say $x=-5$, then $\frac{dx}{dt}=((-5)^{2}-16)^{2}=(25 - 16)^{2}=81>0$, so $x$ is increasing.
In the interval $-40$, so $x$ is increasing.
In the interval $x>4$, say $x = 5$, then $\frac{dx}{dt}=(5^{2}-16)^{2}=81>0$, so $x$ is increasing.
Since solutions move away from $x=-4$ and $x = 4$ as $t$ increases, the differential equation has no stable critical points.

Answer:

B. The differential equation has no stable critical points