Chapter 6, Problem 67EP

$\text{(a)}$

To determine

To find: The proportion of the townspeople have heard the rumor when it is spreading the fastest.

Answer to Problem 67EP

The answer is $\frac{1}{2}$

Explanation of Solution

Given information:

The differential equation is $dy/dt=1.2y(1-y)$ .

Calculation:

The percentage of those who will hear the rumour is all that is required by the textbook; NOT required to actually determine the carrying capacity and solve for half.

  $\frac{1}{2}$ of the population.

Therefore, the required proportion of the townspeople have heard the rumor when it is spreading the fastest is $\frac{1}{2}$ .

$\text{(b)}$

To determine

To find: The $\text{y}$ explicitly as a function of $t$ .

Answer to Problem 67EP

The answer is $y=\frac{1}{1+9e^{-1.2t}}$

Explanation of Solution

Given information:

The differential equation is $dy/dt=1.2y(1-y)$ .

Calculation:

  $\begin{array}{c}\frac{1}{y(1-y)}dy=1.2dt\\ {{\displaystyle \int }}^{\text{}}\frac{1}{y(1-y)}dy={{\displaystyle \int }}^{\text{}}1.2dt\\ \frac{1}{y(1-y)}=\frac{A}{y}+\frac{B}{1-y}\\ =\frac{A(1-y)}{y(1-y)}+\frac{By}{(1-y)y}\\ =\frac{A(1-y)+By}{y(1-y)}\\ A(1-y)+By=\mathrm{1.....}(1)\end{array}$

Put,

  $y=1$ into equation $(1)$

And get:

  $\begin{array}{c}A(1-1)+B\cdot 1=1\\ \to A\cdot 0+B=1\\ \to B=1\end{array}$

Put,

  $y=0$ into equation $(1)$

And get:

  $\begin{array}{c}A(1-0)+B\cdot 0=1\\ \to 1A+B\cdot 0=1\\ \to A=1\end{array}$

Therefore,

  $\frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y}$

Thus,

  $\begin{array}{c}{{\displaystyle \int }}^{\text{}}\frac{1}{y(1-y)}dy={{\displaystyle \int }}^{\text{}}1.2dt\\ \to {{\displaystyle \int }}^{\text{}}\left(\frac{1}{y}+\frac{1}{1-y}\right)dy={{\displaystyle \int }}^{\text{}}1.2dt\\ {{\displaystyle \to \int }}^{\text{}}\frac{1}{y}dy+{{\displaystyle \int }}^{\text{}}\frac{1}{1-y}dy={{\displaystyle \int }}^{\text{}}1.2dt\\ \to \ln \left|y\right|-\ln |1-y|=1.2t+C\\ \to \ln |1-y|-\ln \left|y\right|=-1.2t-C\\ \to \ln \frac{|1-y|}{\left|y\right|}=-1.2t-C\\ \to \frac{1-y}{y}=e^{-1.2t-C}\\ \to \frac{1}{y}-1=e^{-1.2t}\cdot e^{-C}\\ \to \frac{1}{y}=1+e^{-1.2t}\cdot e^{-C}\end{array}$

Put,

  $\begin{array}{c}y=0.1\\ t=0\end{array}$

To find $C$ :

  $\begin{array}{c}\frac{1}{0.1}=1+e^{-1.2\cdot 0}\cdot e^{-C}\\ \to 10=1+e^0\cdot e^{-C}\\ \to 9=e^{-C}\end{array}$

Thus,

  $\begin{array}{c}\frac{1}{y}=1+9e^{-1.2t}\\ \to y=\frac{1}{1+9e^{-1.2t}}\end{array}$

Therefore, the required $\text{y}$ explicitly as a function of $t$ is $y=\frac{1}{1+9e^{-1.2t}}$ .

$(c)$

To determine

To find: The time $t$ is the rumor spreading the fastest.

Answer to Problem 67EP

The answer is: About $1.831$ days

Explanation of Solution

Given information:

The differential equation is $dy/dt=1.2y(1-y)$ .

Calculation:

When half the population of a town is aware of the rumour, it spreads quickly.

  $\begin{array}{c}y=\frac{1}{1+9e^{-1.2t}}\\ \to \frac{1}{2}=\frac{1}{1+9e^{-1.2t}}\\ \to 1+9e^{-1.2t}=2\\ \to e^{-1.2t}=\frac{1}{9}\\ \to \ln e^{-1.2t}=\ln \frac{1}{9}\\ \to -1.2t=\ln \frac{1}{9}\\ \to t=-\frac{\ln 1/9}{1.2}\\ \to t\approx 1.831\text{ days }\end{array}$

Therefore, the required about $1.831$ days is the rumor spreading the fastest.