Final answer:
The equilibrium solutions of the given differential equation are y = 0, y = 2, and y = 1. The equilibrium point y = 0 is stable, the equilibrium point y = 2 is unstable, and the equilibrium point y = 1 is neutral.
Step-by-step explanation:
The autonomous differential equation is given as y′(t) = yy² − 3yy + 2. To find the equilibrium solutions, we set y' equal to zero and solve for y. So we have:
0 = y(y² − 3y + 2)
0 = y(y − 2)(y − 1)
Therefore, the equilibrium solutions are y = 0, y = 2, and y = 1.
To classify the equilibrium solutions as stable or unstable, we need to examine the signs of y' on either side of each equilibrium point. When y = 0, y' = 2, so the equilibrium point y = 0 is stable. When y = 2, y' = -2, so the equilibrium point y = 2 is unstable. When y = 1, y' = 0, so the equilibrium point y = 1 is neutral.