Identify equilibrium values and determine which are stable and which are unstable.
-$$\frac { d y } { d x } = y ^ { 2 } - 3 y$$
A) y = 3 is a stable equilibrium value and y = 0 is an unstable equilibrium.
B) y = 0 is a stable equilibrium value and y = 3 is an unstable equilibrium.
C) y = 0 and y = 3 are unstable equilibrium values.
D) There are no equilibrium values.

Question

Quizplus · Accepted Answer

To find equilibrium values, we set $\frac{dy}{dx}=0$. Solving $y^2-3y=0$ gives us $y=0$ or $y=3$. To determine stability, we can use the derivative test. Taking the derivative of $\frac{dy}{dx}=y^2-3y$, we get $\frac{d^2y}{dx^2}=2y-3$. 
When $y=0$, we have $\frac{d^2y}{dx^2}=-3$, which is negative. This means that $y=0$ is a stable equilibrium. 
When $y=3$, we have $\frac{d^2y}{dx^2}=3$, which is positive. This means that $y=3$ is an unstable equilibrium. 
Therefore, the best choice is B: y = 0 is a stable equilibrium value and y = 3 is an unstable equilibrium.

Identify Equilibrium Values and Determine Which Are Stable and Which dydx=y2−3y\frac { d y } { d x } = y ^ { 2 } - 3 ydxdy​=y2−3y

Identify Equilibrium Values and Determine Which Are Stable and Which $\frac { d y } { d x } = y ^ { 2 } - 3 y$