Identify equilibrium values and determine which are stable and which are unstable.
-$$\frac { d y } { d x } = y ^ { 2 } - 16$$
A) y = 4 is a stable equilibrium value and y = -4 is an unstable equilibrium.
B) y = -4 and y = 5 are stable equilibrium values.
C) y = -4 is a stable equilibrium value and y = 4 is an unstable equilibrium.
D) There are no equilibrium values.

Question

Quizplus · Accepted Answer

To find equilibrium values, we set the derivative equal to zero: $y^2-16=0 \Rightarrow y=\pm4$. To determine the stability of each equilibrium, we can use the second derivative test: $\frac{d^2y}{dx^2}=2y$. When $y=4$, we have $\frac{d^2y}{dx^2}>0$, which means that it is a local minimum and stable. When $y=-4$, we have $\frac{d^2y}{dx^2}<0$, which means that it is a local maximum and unstable. Therefore, choice C is correct.

Identify Equilibrium Values and Determine Which Are Stable and Which dydx=y2−16\frac { d y } { d x } = y ^ { 2 } - 16dxdy​=y2−16

Identify Equilibrium Values and Determine Which Are Stable and Which $\frac { d y } { d x } = y ^ { 2 } - 16$