Determine the equilibrium solutions of y′=y(y−3)(y+2)y' = y(y-3)(y+2)y′=y(y−3)(y+2).
y=0,3,−2y = 0, 3, -2y=0,3,−2
y=1,−1,2y = 1, -1, 2y=1,−1,2
y=0,−3,2y = 0, -3, 2y=0,−3,2
y=3,−2y = 3, -2y=3,−2