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# Find the equilibrium points?

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

### 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.

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Grigori S. | Certified Physics and Math Teacher G.S.Certified Physics and Math Teacher G.S.
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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)  = 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.

Martin S. | Mathematics and Physics Tutor For HireMathematics and Physics Tutor For Hire
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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