Chapter G, Problem 41E

To determine

To calculate: The integral to give an equation relating the female population to time.

Answer to Problem 41E

The integral to give an equation relating the female population to time is $t=-\ln \left|P\right|-\frac{1}{9}\ln \left(P+1000\right)+\ln \left|10000\right|+\frac{1}{9}\ln \left(11000\right)$ .

Explanation of Solution

Given Information:

  $t={\displaystyle \int \frac{P+S}{P\left[\left(r-1\right)P-S\right]}}dP$

Where,

  $P$ = number of female insects

  $S$ = number of sterile males

  $r$ = populations natural growth rate

  $t=$ time

Insect population of females = $10,000$

  $r$ = $0.10$

Sterile males= $900$

Calculation:

The population of $10,000$ females grown at a rate of $r=0.10$

The number of added sterile males are= $900$

Then the integral will be,

  $\begin{array}{l}\Rightarrow t={\displaystyle \int \frac{P+S}{P\left(-0.90P-900\right)}}dP\\ \Rightarrow t={\displaystyle \int \frac{P+S}{-P\left(0.9P-900\right)}}dP\end{array}$

The partial fractions of $\frac{P+900}{-P\left(0.9P+900\right)}$ is,

  $\frac{P+900}{-P\left(0.9P+900\right)}=\frac{A}{0.9P+900}+\frac{B}{P}$

Now, multiply both sides with $-P\left(0.9P+900\right)$

  $\begin{array}{l}\frac{\left(P+900\right)\left(-P\left(0.9P+900\right)\right)}{-P\left(0.9P+900\right)}=\frac{A\left(-P\left(0.9P+900\right)\right)}{P+10}+\frac{B\left(-P\left(0.9P+900\right)\right)}{P}\\ P+900=-A\left(P\right)-B\left(0.9P+900\right)\end{array}$

By using the coverup method, plug in $P=0$ into the equation $P+900=-A\left(P\right)-B\left(0.9P+900\right)$

  $\begin{array}{l}0+900=-A\left(P\right)-B\left(0.9\left(0\right)+900\right)\\ 900=-900B\\ B=-1\end{array}$

Plug in $P=-\frac{900}{0.9}$ into equation $P+900=-A\left(P\right)-B\left(0.9P+900\right)$

  $\begin{array}{l}-\frac{900}{0.9}+900=-A\left(-\frac{900}{0.9}\right)-B\left(0.9\left(-\frac{900}{0.9}\right)+900\right)\\ -100=1000A\\ A=-\frac{100}{1000}\\ A=0.1\end{array}$

The partial fractions of $\frac{P+900}{-P\left(0.9P+900\right)}$ is $-\frac{0.1}{P+10}-\frac{1}{P}$

  $\begin{array}{l}t={\displaystyle \int \left(-\frac{1}{P}-\frac{0.1}{0.9P+900}\right)}dP\\ ={\displaystyle \int \left(-1\right)}\left(\frac{1}{P}\right)dP-{\displaystyle \int \left(0.1\right)\left(\frac{1}{0.9P+900}\right)}dP\\ =-{\displaystyle \int \left(\frac{1}{P}\right)}dP-{\displaystyle \int \frac{1}{9}}\left(\frac{1}{P+1000}\right)dP\\ =-\text{ln}\left|P\right|-\frac{1}{9}\text{In}\left(P+1000\right)+C\end{array}$

The initial condition of $\text{t=0,P=10,000}$ is,

  $\begin{array}{l}0=-\text{ln}\left|10000\right|-\frac{1}{9}\ln \left(10000+1000\right)+C\\ 0=-\ln \left|10000\right|-\frac{1}{9}\ln \left(11000\right)+C\\ C=\ln \left|10000\right|+\frac{1}{9}\ln \left(11000\right)\\ t=-\ln \left|P\right|-\frac{1}{9}\ln \left(P+1000\right)+\ln \left|10000\right|+\frac{1}{9}\ln \left(11000\right)\end{array}$