Chapter 6.5, Problem 28E

To determine

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

Answer to Problem 28E

The slope field:

  $P=\frac{700}{1+69e^{-0.56t}}$

Explanation of Solution

Given information:

The differential equation is: $\frac{dP}{dt}=0.0008P(700-P)$

And $P=10$ when

  $t=0$

Calculation:

  $\begin{array}{c}\frac{1}{P(700-P)}dP=0.0008dt\\ \Rightarrow \int \frac{1}{P(700-P)}dP=\int 0.0008dt\\ \frac{1}{P(700-P)}=\frac{A}{P}+\frac{B}{700-P}\\ =\frac{A(700-P)}{P(700-P)}+\frac{BP}{(700-P)P}\\ =\frac{A(700-P)+BP}{P(700-P)}\end{array}$

So it follows that,

  $A(700-P)+BP=\mathrm{1.....}(1)$

Put, $P=700$ into the Equation $(1)$ and get:

  $\begin{array}{c}A(700-700)+B\cdot 700=1\\ \Rightarrow A\cdot 0+700B=1\\ \Rightarrow 700B=1\\ \Rightarrow B=\frac{1}{700}\end{array}$

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

  $\begin{array}{c}A(700-0)+B\cdot 0=1\\ \Rightarrow 700A+B\cdot 0=1\\ \Rightarrow 700A=1\\ \Rightarrow A=\frac{1}{700}\end{array}$

Therefore, $\frac{1}{P(700-P)}=\frac{1}{700P}+\frac{1}{700(700-P)}$ , thus

  $\begin{array}{c}\int \frac{1}{700P}dP+\int \frac{1}{700(700-P)}dP=\int 0.0008dt\\ \Rightarrow \frac{1}{700}\int \frac{1}{P}dP+\frac{1}{700}\int \frac{1}{700-P}dP=0.0008\int dt\\ \Rightarrow \frac{1}{700}\ln \left|P\right|-\frac{1}{700}\ln |700-P|=0.0008t+C\\ \Rightarrow \ln |700-P|-\ln \left|P\right|=-0.56t-C\\ \Rightarrow \ln \frac{|700-P|}{\left|P\right|}=-0.56t-C\\ \Rightarrow \frac{700-P}{P}=e^{-0.56t-C}\\ \Rightarrow \frac{700}{P}-1=e^{-0.56t}\cdot e^{-C}\\ \Rightarrow \frac{700}{P}=1+e^{-0.56t}\cdot e^{-C}\end{array}$

Put $P=10$ and

  $t=0$ to find $C$ :

  $\begin{array}{c}\frac{700}{10}=1+e^{-0.56\cdot 0}\cdot e^{-C}\\ \Rightarrow 70=1+e^0\cdot e^{-C}\\ \Rightarrow 69=e^{-C}\end{array}$

Thus,

  $\begin{array}{c}\frac{700}{P}=1+69e^{-0.56t}\\ \Rightarrow P=\frac{700}{1+69e^{-0.56t}}\end{array}$

  

Therefore the required slope field for the differential equation $\frac{dP}{dt}=0.0008P(700-P)$ is $P=\frac{700}{1+69e^{-0.56t}}$ .