Chapter 6.5, Problem 29E

To determine

To find : The slope field for the differential equation and verify that.

Answer to Problem 29E

The slope field:

  $P=\frac{1200}{1+59e^{-0.24t}}$

Explanation of Solution

Given information:

The differential equation is: $\frac{dP}{dt}=0.0002P(1200-P)$

And $P=20$ when

  $t=0$

Calculation:

  $\begin{array}{c}\frac{1}{P(1200-P)}dP=0.0002dt\\ \int \frac{1}{P(1200-P)}dP=\int 0.0002dt\\ \frac{1}{P(1200-P)}=\frac{A}{P}+\frac{B}{1200-P}\\ =\frac{A(1200-P)}{P(1200-P)}+\frac{BP}{(1200-P)P}\\ =\frac{A(1200-P)+BP}{P(1200-P)}\end{array}$

So it follows that,

  $A(1200-P)+BP=\mathrm{1.....}(1)$

Put $P=1200$ into the Equation $(1)$ and get:

  $\begin{array}{c}A(1200-1200)+B\cdot 1200=1\\ \Rightarrow A\cdot 0+1200B=1\\ \Rightarrow 1200B=1\\ \Rightarrow B=\frac{1}{1200}\end{array}$

Put $P=0$ into the Equation $(1)$ and get:

  $\begin{array}{c}A(1200-0)+B\cdot 0=1\\ \Rightarrow 1200A+B\cdot 0=1\\ \Rightarrow 1200A=1\\ \Rightarrow A=\frac{1}{1200}\end{array}$

Therefore, $\frac{1}{P(1200-P)}=\frac{1}{1200P}+\frac{1}{1200(1200-P)}$ . Thus

  $\begin{array}{c}\int \frac{1}{P(1200-P)}dP=\int 0.0002dt\\ \Rightarrow \int \left(\frac{1}{1200P}+\frac{1}{1200(1200-P)}\right)dP=\int 0.0002dt\\ \Rightarrow \int \frac{1}{1200P}dP+\int \frac{1}{1200(1200-P)}dP=\int 0.0002dt\\ \Rightarrow \frac{1}{1200}\int \frac{1}{P}dP+\frac{1}{1200}\int \frac{1}{1200-P}dP=0.0002\int dt\\ \Rightarrow \frac{1}{1200}\ln \left|P\right|-\frac{1}{1200}\ln |1200-P|=0.0002t+C\\ \Rightarrow \ln |1200-P|-\ln \left|P\right|=-0.24t-C\\ \Rightarrow \ln \frac{|1200-P|}{\left|P\right|}=-0.24t-C\\ \Rightarrow \frac{1200-P}{P}=e^{-0.24t-C}\\ \Rightarrow \frac{1200}{P}-1=e^{-0.24t}\cdot e^{-C}\\ \Rightarrow \frac{1200}{P}=1+e^{-0.24t}\cdot e^{-C}\end{array}$

Put $P=20$ and

  $t=0$ to find $C$ :

  $\begin{array}{c}\frac{1200}{20}=1+e^{-0.24\cdot 0}\cdot e^{-C}\\ \Rightarrow 60=1+e^0\cdot e^{-C}\\ \Rightarrow 59=e^{-C}\end{array}$

Thus,

  $\begin{array}{c}\frac{1200}{P}=1+59e^{-0.24t}\\ \Rightarrow P=\frac{1200}{1+59e^{-0.24t}}\end{array}$

  

Therefore the required slope field for the differential equation $\frac{dP}{dt}=0.0002P(1200-P)$ is $P=\frac{1200}{1+59e^{-0.24t}}$