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
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

Explanation:

Step1: Find critical points

Set $\frac{dx}{dt}=(x^{2}-16)^{2}=0$. Then $x^{2}-16 = 0$, so $x=\pm4$.

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 a critical - point $x_0$, if $f^\prime(x_0)<0$, the critical point is stable; if $f^\prime(x_0)>0$, the critical point is unstable; if $f^\prime(x_0)=0$, further analysis is needed.
When $x = 4$, $f^\prime(4)=4\times4\times(4 - 4)\times(4 + 4)=0$. When $x=-4$, $f^\prime(-4)=4\times(-4)\times(-4 - 4)\times(-4 + 4)=0$.
We can also use the fact that $\frac{dx}{dt}=(x^{2}-16)^{2}\geq0$ for all $x$. Near $x = 4$ and $x=-4$, $\frac{dx}{dt}$ does not change sign from positive to negative. So there are no stable critical points.
For unstable non - semistable points, since $\frac{dx}{dt}=(x^{2}-16)^{2}\geq0$ and the derivative of $\frac{dx}{dt}$ with respect to $x$ is zero at $x=\pm4$, there are no unstable non - semistable points.

Answer:

B. The differential equation has no stable critical points
B. The differential equation has no unstable critical points that are not semistable