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

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)}.