Question

Suppose that news spreads through a city of fixed size of 700000 people at a time rate proportional to the number of people who have not heard the news.

(a) Formulate a differential equation and initial condition for y(t), the number of people who have heard the news t days after it has happened.
No one has heard the news at first, so y(0)=0. The "time rate of increase in the number of people who have heard the news is proportional to the number of people who have not heard the news" translates into the differential equation
\frac{dy}{dt} = k (),where k is the proportionality constant.
(b) 9 days after a scandal in City Hall was reported, a poll showed that 350000 people have heard the news. Using this information and the differential equation, solve for the number of people who have heard the news after t days.
y(t) =

Answer 1

(a)
`(dy)/(dt)=k\left ( 700000-y(t) \right )`

(b)
`y(t)=700000-700000e^{(1)/(9)\ln \left ( (1)/(2) \right )t}`

Explanation:

Given: News spreads through a city of fixed size of 700000 people at a time rate proportional to the number of people who have not heard the news.

To find:

(a) a differential equation

(b) number of people who have heard the news after t days

Solution:

(a)

Total number of people in a city = 700000

As y(t) denotes the number of people who have heard the news t days after it has happened

Number of people who have not heard the news = 700000 - y(t)

So, differential equation is
`(dy)/(dt)=k\left ( 700000-y(t) \right )`

Here, k is the proportionality constant.

(b)

Integrate both sides of the differential equation.

`(dy)/(dt)=k\left ( 700000-y(t) \right )\\\int (dy)/(\left ( 700000-y(t) \right ))=\int k\,dt\\ln\left ( 700000-y \right )=kt+C\,\,\left \{ \because \int (dy)/(y)=\ln y+C \right \}`

Use
`y(0)=0`

`ln\left ( 700000-y \right )=kt+C\\\ln (700000)=C\\\Rightarrow ln\left ( 700000-y \right )=kt+\ln (700000)`

As a poll showed that 350000 people have heard the news 9 days after a scandal in City Hall was reported,
`y(9)=350000`

`ln\left ( 700000-y \right )=kt+\ln (70000)\\ln\left ( 700000-350000 \right )=kt+\ln (700000)\\\ln (350000)=9k+\ln (700000)\\9k=\ln (350000)-\ln (700000)\\`

`9k=\ln \left ( (350000)/(700000) \right )\\9k=\ln \left ( (1)/(2) \right )\\k=(1)/(9)\ln \left ( (1)/(2) \right )\\ln\left ( 700000-y \right )=(1)/(9)\ln \left ( (1)/(2) \right )t+\ln (700000)\\ln\left ( 700000-y \right )-\ln (700000)=(1)/(9)\ln \left ( (1)/(2) \right )t\\`

`\ln \left ( (700000-y )/(700000) \right )=(1)/(9)\ln \left ( (1)/(2) \right )t\\ (700000-y )/(700000)=e^{(1)/(9)\ln \left ( (1)/(2) \right )t}\\700000-y=700000e^{(1)/(9)\ln \left ( (1)/(2) \right )t}\\y=700000-700000e^{(1)/(9)\ln \left ( (1)/(2) \right )t}\\`