Identify equilibrium values and determine which are stable and which are unstable.
-$$\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { y } + 5 ) ( \mathrm { y } + 4 )$$
A) y = 4 is a stable equilibrium value and y = -5 is an unstable equilibrium.
B) y = -5 is a stable equilibrium value and y = 4 is an unstable equilibrium.
C) y = -4 is a stable equilibrium value and y = 5 is an unstable equilibrium.
D) y = 5 is a stable equilibrium value and y = 4 is an unstable equilibrium.

Question

Quizplus · Accepted Answer

To determine the stability of the equilibrium points, we first identify them by setting the derivative equal to zero: $(y + 5)(y + 4) = 0$, which gives $y = -5$ and $y = -4$ as the roots. However, the correct roots from the given equation are $y = -5$ and $y = -4$, but since there's a mistake in identifying the equilibrium values in the options, we proceed with the provided values in the options, assuming a typographical error in the explanation phase. The correct equilibrium points from the given equation are $y = -5$ and $y = -4$, but based on the options, we consider $y = -5$ and $y = 4$. To determine stability, we look at the sign of the derivative on either side of these points. For $y = -5$, the derivative changes from negative to positive as y increases through -5, indicating a stable equilibrium. For $y = 4$, the scenario described doesn't match the given equation's dynamics, suggesting a misinterpretation in the options provided. The correct approach involves analyzing the sign of the derivative around the actual equilibrium points, $y = -5$ and $y = -4$, to determine their stability correctly. Given the misalignment in the options, the closest match to the expected analysis is option B, assuming it refers to the correct equilibrium points $y = -5$ and $y = -4$ and their stability, despite the typographical error in stating $y = 4$ instead of the correct $y = -4$. This explanation acknowledges the discrepancy and aims to clarify the intended analysis process.

Identify Equilibrium Values and Determine Which Are Stable and Which dydx=(y+5)(y+4)\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { y } + 5 ) ( \mathrm { y } + 4 )dxdy​=(y+5)(y+4)

Identify Equilibrium Values and Determine Which Are Stable and Which $\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { y } + 5 ) ( \mathrm { y } + 4 )$