df/dt = bf × (af – 1), with initial condition f(0) = 1, solve equation for f(t)

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

ft/dt = bf * (af-1)

We know that:

df/dt = f'(t)

Then, f(t)= integral f'(t)

==> f(t) = intg (bf *(af-1) dt

              = intg (abf^2 - bf) dt

               = abf^3/3 - bf^2 /2 + C

               = (ab/3)f^3 - (b/2)f^2 + C

But: f(0) = 1

==> f(0) = 0 - 0 + C = 1

==> C = 1

==> f(t) = (ab/3)f^3 - (b/2)f^2 + 1

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

df/dt = bf*(af-1).  the initial condition isf(o) = 1.

To solve the differential equation  we rewrite the equation as:

We rewrite the given equation as below:

dt = df/bf(af-1).

dt = df{ 1/2b(af-1) -1/2abf}.

Now we integrate .

t  = (1/2b){ln(af-1)}/a - (1/2ab)lnf + a constant of integration.

t = (1/2ab){ln(af-1)/f} + C

Therefore ln{(af-1)/f} = 2ab(t-C).

Taking antilogarithms, we get:

(af-1)/f = e^2ab(t-C).

af -1 = f*e^2ab(t-C)

af - fe^2ab(t-c) = 1

f [a-e^2ab(t-c)] = 1

f = 1/[a-e^2ab(t-C)]......(1)

Now we determine C using the given initial condition f(0) = 1.

f(0) = 1 implies  1/[a-e^2ab(0-C)] = 1

1/[a-e^2abC] = 1

1 = a-e^2abC

e^2abC = a-1

Taking logarithms, we get:

2abC = ln(a-1)

C = (1/2ab)ln(a-1).

Therefore from (1) we get: f = f(t) = 1/[a-e^2ab(t- (1/2ab)ln(a-1)]

f(t) = 1/[a- (e^2abt)/(a-1)^(1/2ab)].

f(t) = {(a-1)^(1/2ab)}/{(a-1)^(1/2ab) - e^(2abt)}.

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