Find the equilibrium points of dy/dt=ay+by^2, a>0, b>0, yo≥0.

## Find the equilibrium points?

# 2 Answers

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)

or

(y/(a+by))^{ 1/a} = C e
^{t (4)}

Raise both sides into "a" power and you will obtain

y/(a+by) = C^{a} e^{at} = C_{1} e
^{at } (5)

(C^{a} is replaced by an arbitrary cxnstant C_{1. }Taking into account the initial condition y(0) = y_{0}

we can find that

y_{0} = C_{1}a/(1-C_{1}b) (6)

Substituiting (6) into (5) and solving for "y" we will obtain the explicit expression for y(t):

y(t) = a y_{0}e^{at}/[a +by_{0}(1-e^{at})] (7)

If you calculate the derivative dy/dt you will find that it equals zero if

a+by_{0} = 0 or y_{0} = - a/b

Thus, the intial state (t = 0, y_{0} = -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.

The equilibrium points are at the values where dy/dt = 0.

Simply factor the y out of (ay + by^2) to get y(a + by).

ay + by^2 = y(a + by) = 0

a + by = 0 --> a = -by --> y = -a/b

y = 0, -a/b

## Comments

Grigori, so the answer is y=0, -a/b, right?

Yes. I just wanted to say that physically attainable point is y = -a/b, despite formally both solutions are correct.

Yes. I just wanted to say that physically attainable point is y = -a/b, despite formally both solutions are correct.

Comment