Find the stability of this differential equation

Question

Find the stability of this differential equation and explain if it's stable or unstable

 

dy/dt = (1-y)(y+3)

Answer 1

Set dy/dt = 0 to find the equilibrium points:

 

0 = (1 - y)(y + 3)

 

y = 1 or y = -3

 

The equilibrium point is stable if values near it approach it or unstable if values near it move away as t → ∞.

 

y = 1: dy/dt = (1 - y)(y + 3) is positive if y < 1 and negative if y > 1 near the neighborhood of y = 1.  The values are attracted to y = 1, so the equilibrium point is stable.

 

y = -3: dy/dt = (1 - y)(y + 3) is negative if y < -3 and positive if y > -3 near the neighborhood of y = -3.  The values are repelled by y = -3, so the equilibrium point is unstable.