Identify equilibrium values and determine which are stable and which are unstable.
-$$\frac { d y } { d x } = y ^ { 2 } - 5 y$$
A) y = 0 and y = 5 are unstable equilibrium values.
B) y = 0 is a stable equilibrium value and y = 5 is an unstable equilibrium.
C) y = 5 is a stable equilibrium value and y = 0 is an unstable equilibrium.
D) There are no equilibrium values.

Question

Quizplus · Accepted Answer

The equilibrium values are found by setting the derivative to zero: $y^2 - 5y = 0$, which gives $y(y-5) = 0$. Thus, $y = 0$ and $y = 5$ are equilibrium values. To determine stability, consider the sign of $\frac{dy}{dx}$ around these points. For $y = 0$, $\frac{dy}{dx} < 0$ for $y < 0$ and $\frac{dy}{dx} > 0$ for $0 < y < 5$, indicating stability. For $y = 5$, $\frac{dy}{dx} > 0$ for $y > 5$ and $\frac{dy}{dx} < 0$ for $0 < y < 5$, indicating instability.

Identify Equilibrium Values and Determine Which Are Stable and Which dydx=y2−5y\frac { d y } { d x } = y ^ { 2 } - 5 ydxdy​=y2−5y

Identify Equilibrium Values and Determine Which Are Stable and Which $\frac { d y } { d x } = y ^ { 2 } - 5 y$