Identify equilibrium values and determine which are stable and which are unstable.
-$$y ^ { \prime } = ( y - 1 ) ( y - 3 ) ( y - 4 )$$
A) y = 5, y = 1 and y = 4 are unstable equilibria.
B) y = 1 is a stable equilibrium value and y = 3 and y = 4 are unstable equilibria.
C) y = 4 is a stable equilibrium value and y = 3 and y = 1 are unstable equilibria.
D) y = 3 is a stable equilibrium value and y = 1 and y = 4 are unstable equilibria.

Question

Quizplus · Accepted Answer

The equilibrium values are found by setting $y' = 0$ and solving for $y$: $$(y-1)(y-3)(y-4) = 0.$$ This gives three equilibrium values: $y=1, y=3,$ and $y=4$. To determine stability, we can use the first derivative test. An equilibrium value is stable if $y'<0$ to the left of it and $y'>0$ to the right of it. Conversely, an equilibrium value is unstable if $y'<0$ to the right of it and $y'>0$ to the left of it. For $y<1$, we have $y-1<0$, $y-3<0$, and $y-4<0$, so $y'<0$. Between $1$ and $3$, we have $y-1<0$, $y-3<0$, and $y-4>0$, so $y'>0$. Thus, $y=1$ is a stable equilibrium. Between $3$ and $4$, we have $y-1<0$, $y-3>0$, and $y-4>0$, so $y'<0$. Thus, $y=3$ is a stable equilibrium. For $y>4$, we have $y-1>0$, $y-3>0$, and $y-4>0$, so $y'>0$. Therefore, the equilibrium values $y=1$ and $y=3$ are stable, and the equilibrium value $y=4$ is unstable. Therefore, answer choice D is correct.

Identify Equilibrium Values and Determine Which Are Stable and Which y′=(y−1)(y−3)(y−4)y ^ { \prime } = ( y - 1 ) ( y - 3 ) ( y - 4 )y′=(y−1)(y−3)(y−4)

Identify Equilibrium Values and Determine Which Are Stable and Which $y ^ { \prime } = ( y - 1 ) ( y - 3 ) ( y - 4 )$