Chapter 7.5, Problem 27E

To determine

To solve the initial value problem using partial fraction. And verify that the solution conforms to the slope field.

Answer to Problem 27E

The required solution is $P=\frac{200}{1+24e^{-1.2t}}$ .

Explanation of Solution

Given information: The function,

  $\frac{dp}{dt}=0.0006P\left(200-P\right)$ and $P=8$ when $t=0$ .

Calculation: The required solution is obtained as,

  $\begin{array}{l}\frac{dp}{dt}=0.0006P\left(200-P\right)\\ \frac{1}{P\left(200-P\right)}dP=0.0006\\ {\displaystyle \int \frac{1}{P\left(200-P\right)}dP={\displaystyle \int 0.0006dt}}\end{array}$

To integrate the left side we must rewrite it as two fraction,

  $\begin{array}{l}\frac{1}{P\left(200-P\right)}=\frac{A}{P}+\frac{B}{200-P}\\ 1=A\left(200-P\right)+B\left(P\right)\end{array}$

When P = 0 then,

  $1=A\left(200-0\right)+B\left(0\right)\to 1=200A\to A=\frac{1}{200}$

When P = 200 then,

  $1=A\left(200-200\right)+B\left(200\right)\to 1=200B\to B=\frac{1}{200}$

This then gives,

  $\begin{array}{l}\frac{1}{P\left(200-P\right)}=\frac{\frac{1}{200}}{P}+\frac{\frac{1}{200}}{200-P}\\ \frac{1}{P\left(200-P\right)}=\frac{\frac{1}{200}}{P}+\frac{\frac{1}{200}}{200-P}\\ =\frac{\frac{1}{200}}{P}+\frac{\frac{1}{200}}{-\left(P-200\right)}\\ =\frac{\frac{1}{200}}{P}-\frac{\frac{1}{200}}{\left(P-200\right)}\end{array}$

Substitute the partial fraction,

  $\begin{array}{l}\ln \left|P\right|-\ln \left|P-200\right|=1.2t+C\\ \ln \left|\frac{P}{P-200}\right|=1.2t+C\\ \frac{P}{P-200}=e^{1.2t+C}\\ \frac{P}{P-200}=e^{1.2t}\cdot e^C\\ \frac{P}{P-200}=C{e}^{1.2t}\end{array}$

Substitute in P = 8 and t = 0 to find C,

  $\begin{array}{l}\frac{8}{8-200}=C{e}^0\\ \frac{8}{-192}=C\left(1\right)\\ C=-\frac{1}{24}\end{array}$

Substitute $C=-\frac{1}{24}$ in equation $\frac{P}{P-200}=C{e}^{1.2t}$ ,

  $\begin{array}{l}\frac{P}{P-200}=-\frac{1}{24}{e}^{1.2t}\\ P=-\frac{1}{24}{e}^{1.2t}\left(P-200\right)\\ P=-\frac{1}{24}{e}^{1.2t}P+\frac{25}{3}{e}^{1.2t}\\ \frac{1}{24}{e}^{1.2t}P+P=\frac{25}{3}{e}^{1.2t}\\ P\left(\frac{1}{24}{e}^{1.2t}+1\right)=\frac{25}{3}{e}^{1.2t}\\ P=\frac{\frac{25}{3}{e}^{1.2t}}{\frac{1}{24}{e}^{1.2t}+1}\\ P=\frac{200}{1+24e^{-1.2t}}\end{array}$

Hence the required solution is $P=\frac{200}{1+24e^{-1.2t}}$ .

And the slope of P match with the slopes of the line segments of the slope field.