Chapter G, Problem 40E

To determine

To calculate:

The integral value of the female population to time:

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

Answer to Problem 40E

The integral value of $t={\displaystyle \int \frac{P+S}{P\left[\left(r-1\right)P-S\right]}dP}$ is $t=-\ln P-\frac{1}{9}\left(P+1000\right)+C$ .

Explanation of Solution

Given information:

The integral function:

  $\begin{array}{l}t={\displaystyle \int \frac{P+S}{P\left[\left(r-1\right)P-S\right]}dP}\\ s=900\\ r=0.1\end{array}$

Calculation:

The integral function is given that:

  $\begin{array}{l}t={\displaystyle \int \frac{P+900}{P\left[\left(-0.9\right)P-900\right]}dP}\\ t=-\frac{10}{9}{\displaystyle \int \frac{P+900}{P\left[P+1000\right]}dP}\end{array}$

Now, apply the partial fraction expansion to expand the integrand.

  $\begin{array}{l}\frac{P+900}{P\left[P+1000\right]}=\frac{A}{P}+\frac{B}{P+1000}\\ P+900=A\left(P+1000\right)+BP\\ P+900=P\left(A+B\right)+1000A\\ 1=A+B\\ 900=1000A\\ A=\frac{9}{10}\\ B=\frac{1}{10}\end{array}$

Now can solve for this integral

  $\begin{array}{l}t=-\frac{10}{9}{\displaystyle \int \left(\frac{\frac{9}{10}}{P}+\frac{\frac{1}{10}}{P+1000}\right)dP}\\ t=-\frac{10}{9}\left(\frac{9}{10}\ln P+\frac{1}{10}\ln \left(P+1000\right)\right)+c\\ t=-\ln P-\frac{1}{9}\left(P+1000\right)+c\end{array}$