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

Therefore ,
`y= (Ae^(kt))/(1+Ae^(kt) )`

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 (dy)/(dt)=ky(1-y)`

`\Rightarrow (dy)/(y(1-y))=k \ dt`

Integrating both sides

`\Rightarrow \int(dy)/(y(1-y))=\int k \ dt`

`\Rightarrow \int (dy)/(y)+\int(dy)/(1-y)=\int k \ dt`
`[\because (1)/(y(1-y)=(1)/(y)+(1)/(1-y) ]`

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

`\Rightarrow ln|(y)/(1-y)|=kt +c`

`\Rightarrow (y)/(1-y)= e^(kt+c)`

`\Rightarrow (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= (Ae^(kt))/(1+Ae^(kt) )`

Therefore ,
`y= (Ae^(kt))/(1+Ae^(kt) )`