Question

One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by y. (Use k for the constant of proportionality.)

Answer 1

Answer:

Therefore , $y= \frac{Ae^{kt}}{1+Ae^{kt} }$

Step-by-step explanation:

The fraction of population who have heard rumor = y

The  fraction of population who haven't heard rumor = 1-y

The rate of of spread (y'(t)) is proportional to the product of the fraction of population who have heard rumor the fraction of population who have heard rumor.

Therefore

y'(t) ∝ y (1-y)

⇒ y'(t) =k y (1-y)   [ k = constant of proportional]

$\Rightarrow \frac{dy}{dt}=ky(1-y)$

$\Rightarrow \frac{dy}{y(1-y)}=k \ dt$

Integrating both sides

$\Rightarrow \int\frac{dy}{y(1-y)}=\int k \ dt$

$\Rightarrow \int \frac{dy}{y}+\int\frac{dy}{1-y}=\int k \ dt$              $[\because \frac{1}{y(1-y}=\frac{1}{y}+\frac{1}{1-y} ]$

$\Rightarrow ln \ y- ln \ |1-y| = kt +c$      [ c = arbitrary constant]

$\Rightarrow ln|\frac{y}{1-y}|=kt +c$

$\Rightarrow \frac{y}{1-y}= e^{kt+c}$

$\Rightarrow \frac{y}{1-y}= Ae^{kt}$      [ Here $e^c=A$ ]

$\Rightarrow y = Ae^{kt}(1-y)$

$\Rightarrow y = Ae^{kt}-y Ae^{kt}$

$\Rightarrow y+y Ae^{kt} = Ae^{kt}$

$\Rightarrow y(1+Ae^{kt} )= Ae^{kt}$

$\Rightarrow y= \frac{Ae^{kt}}{1+Ae^{kt} }$

Therefore , $y= \frac{Ae^{kt}}{1+Ae^{kt} }$