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.
$\frac{dx}{dt}=(x^{2}-16)^{2}$

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.

identify any unstable critical point(s) that is/are not semistable. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the differential equation has unstable critical point(s) that is/are not semistable at x =
(simplify your answer. use a comma to separate answers as needed.)
b. the differential equation has no unstable critical points that are not semistable.

identify any semistable 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 semistable critical point(s) at x =
(simplify your answer. use a comma to separate answers as needed.)
b. the differential equation has no semistable critical points

Explanation:

Step1: Find critical points

Set $\frac{dx}{dt}=(x^{2}-16)^{2}=0$. Then $x^{2}-16 = 0$, so $(x - 4)(x + 4)=0$. The critical points are $x=-4,4$.

Step2: Analyze stability

Let $f(x)=(x^{2}-16)^{2}=x^{4}-32x^{2}+256$. Take the derivative $f^\prime(x)=4x^{3}-64x = 4x(x^{2}-16)=4x(x - 4)(x + 4)$.
For $x=-4$ and $x = 4$, $f^\prime(x)=0$. Since $\frac{dx}{dt}=(x^{2}-16)^{2}\geq0$ for all $x$, solutions on either side of $x=-4$ and $x = 4$ move away or stay the same. These are semistable critical - points.

Answer:

1. B. The differential equation has no stable critical points
2. B. The differential equation has no unstable critical points that are not semistable
3. A. The differential equation has semistable critical point(s) at $x=-4,4$