Chapter 7.5, Problem 9E

a.

To determine

To write : a differential equation that is satisfied by $y$

Answer to Problem 9E

The differential equation is $\frac{dy}{dt}=ky\left(1-y\right)$

Explanation of Solution

Given information : The question is

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

Calculation :

  $\frac{dy}{dt}$ is the rate of change

  $y$ is the fraction who have heard the rumor

  $1-y$ is the fraction who have not heard the rumor

  $\frac{dy}{dt}=ky\left(1-y\right)$

Since the rate of change is directly proportional to the fraction who have heard the rumor and not heard the rumor

These function are multiplied together with a constant $k$

b.

To determine

To solve : the differential equation

Answer to Problem 9E

The solution of differential equation is $y=\frac{y_0}{y_0+\left(1-y_0\right)e^{-kt}}$

Explanation of Solution

Given information : The question is

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

Calculation : the solution is

  $\begin{array}{l}\frac{dy}{dt}=ky\left(1-y\right)\\ \frac{dy}{y\left(1-y\right)}=kdt\end{array}$

Re write both sides as integral

  $\begin{array}{l}{\displaystyle \int \frac{dy}{y\left(1-y\right)}={\displaystyle \int kdt}}\\ \frac{1}{y\left(1-y\right)}=\frac{A}{y}+\frac{B}{1-y}\\ A\left(1-y\right)+By=A-Ay+By=1\\ A=1\\ B-A=0\\ B-1=0\\ B=1\\ {\displaystyle \int \frac{1}{y}}dy+{\displaystyle \int \frac{1}{1-y}={\displaystyle \int kdt}}\\ \ln \left|y\right|-\ln \left|1-y\right|=kt+C\\ \ln \left|\frac{y}{1-y}\right|=kt+C\\ e^{\ln \left|\frac{y}{1-y}\right|}=e^{kt+C}\\ \frac{y}{1-y}=A{e}^{kt}\\ y=\left(1-y\right)A{e}^{kt}\\ y=A{e}^{kt}-A{e}^{kt}y\\ y+A{e}^{kt}y=A{e}^{kt}\\ y\left(1+A{e}^{kt}\right)=A{e}^{kt}\\ y=\frac{A{e}^{kt}}{1+A{e}^{kt}}\\ \frac{y}{1-y}=A{e}^{kt}\\ \frac{y_0}{1-y_0}=A\\ y=\frac{\frac{y_0}{1-y_0}}{1+\frac{y_0}{1-y_0}{e}^{kt}}\\ \text{multiply by }\frac{e^{-kt}}{e^{-kt}}\\ y=\frac{y_0}{y_0+\left(1-y_0\right)e^{-kt}}\end{array}$

c.

To determine

To calculate : at what time $90\%$ of the population have heard the rumor

Answer to Problem 9E

At $3:36PM$

  $90\%$ of the population have heard the rumor

Explanation of Solution

Given information : The question is

A small town has $1000$ inhabitants. At $8am$

  $80$ people have heard the rumor. By the noon half of the town has heard

Calculation : let $t=0$ at $8am$ $\begin{array}{l}\frac{80}{100}=0.8\\ \left(0,0.08\right)\left(4,0.5\right)\end{array}$

The second point occurs at $12PM$ , $4$ hours after $8am$

  $\begin{array}{l}0.08=\frac{y_0}{y_0+\left(1-y_0\right)}\\ y=0.08\\ t=0\\ 0.08=y_0\\ 0.5=\frac{0.08}{0.08+0.92e^{-4k}}\\ y=0.5\\ t=4\\ 0.08+0.92e^{-4k}=\frac{0.08}{0.5}=0.16\\ e^{-4k}=\frac{2}{23}\\ \ln e^{-4k}=\ln \left(\frac{2}{23}\right)\\ -4k=\ln \left(\frac{2}{23}\right)\\ k=\frac{-\ln \left(\frac{2}{23}\right)}{4}=0.611\\ y=\frac{0.08}{0.08+0.92e^{-0.611t}}\\ 0.08+0.92e^{-0.611t}=\frac{0.08}{0.9}=\frac{4}{45}\\ e^{-0.611t}=\frac{2}{207}\\ \ln e^{-0.611t}=\ln \left(\frac{2}{207}\right)\\ t=\frac{-\ln \left(\frac{2}{207}\right)}{0.611}=7.593\\ .593hr\times \left(\frac{60\min }{1hr}\right)=35.58\min \\ 8:AM+7:36=3:36PM\end{array}$