Find the equilibrium points of dy/dt=ay+by^2, a>0, b>0, yo≥0.
You have to show the values of "t" that make the derivative dy/dt = 0. Let's find the explicit solution of the equation. After separation of variables we can rewrite it in the following way:
dy/y(a+by) = dt (1)
The rational expression can be written as
1/y(a + by) = (1/a)[(1/y) - b/(a+by)] (2)
Integration gives us
(1/a)[lny - ln(a+by)] = ln C + t (C = const) (3)
(y/(a+by)) 1/a = C e t (4)
Raise both sides into "a" power and you will obtain
y/(a+by) = Ca eat = C1 e at (5)
(Ca is replaced by an arbitrary cxnstant C1. Taking into account the initial condition y(0) = y0
we can find that
y0 = C1a/(1-C1b) (6)
Substituiting (6) into (5) and solving for "y" we will obtain the explicit expression for y(t):
y(t) = a y0eat/[a +by0(1-eat)] (7)
If you calculate the derivative dy/dt you will find that it equals zero if
a+by0 = 0 or y0 = - a/b
Thus, the intial state (t = 0, y0 = -a/b) is a state of equilibrium dy/dt = 0.
Despite dy/dt = 0 if y = 0, this solution has to excluded because, as one can see from (7), y = 0 for
t = - ∞ (not a reachable state). In the meantime, for t = ∞ y = -a/b and dy/dt = 0.