Question

Identify equilibrium values and determine which are stable and which are unstable. dy/dx = y^2 - 1 A. y = -1 is a stable equilibrium value and y = 1 is an unstable equilibrium value. B. y = -1 and y = 1 are stable equilibrium values. C. y = 1 is a stable equilibrium value and y = -1 is an unstable equilibrium value. D. There are no equilibrium values.

          Identify equilibrium values and determine which are stable and which are unstable.

dy/dx = y^2 - 1

A. y = -1 is a stable equilibrium value and y = 1 is an unstable equilibrium value.
B. y = -1 and y = 1 are stable equilibrium values.
C. y = 1 is a stable equilibrium value and y = -1 is an unstable equilibrium value.
D. There are no equilibrium values.

Identify equilibrium values and determine which are stable and which are unstable.

dy/dx = y^2 - 1

A. y = -1 is a stable equilibrium value and y = 1 is an unstable equilibrium value.
B. y = -1 and y = 1 are stable equilibrium values.
C. y = 1 is a stable equilibrium value and y = -1 is an unstable equilibrium value.
D. There are no equilibrium values.

Added by Cynthia J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

05/23/2023

Step 1

These are the values of y for which dy/dx = 0. So, we have: $$ \frac{dy}{dx} = y - 1 = 0 $$ Solving for y, we get: $$ y = 1 $$ So, there is only one equilibrium value, which is $y = 1$. Now, we need to determine the stability of this equilibrium value. To do Show more…

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Identify equilibrium values and determine which are stable and which are unstable. dy/dx = y^2 - 1 A. y = -1 is a stable equilibrium value and y = 1 is an unstable equilibrium value. B. y = -1 and y = 1 are stable equilibrium values. C. y = 1 is a stable equilibrium value and y = -1 is an unstable equilibrium value. D. There are no equilibrium values.