Chapter 7.5, Problem 30E

To determine

To solve the initial value problem and generate a slope field for the differential equation and verify whether the solution conforms to the slope field.

Answer to Problem 30E

The solution of differential equation is $P=\frac{5000}{99e^{-0.05t}+1}$ .

Explanation of Solution

Given information:

The differential equation is

  $\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)\text{ and }P=50\text{ when }t=0$

Calculation:

Since, $\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)$ using separation of variable method.

  $\begin{array}{l}\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)\\ \frac{1}{P\left(5000-P\right)}dP={10}^{-5}dt\\ \text{integrating both side}\\ {\displaystyle \int \frac{1}{P\left(5000-P\right)}dP={\displaystyle \int {10}^{-5}dt}}\\ \text{Using partial fraction}\end{array}$

  $\begin{array}{l}\because \frac{1}{P\left(5000-P\right)}=\frac{A}{P}+\frac{B}{5000-P}=\frac{A\left(5000-P\right)+B\left(P\right)}{P\left(5000-P\right)}\\ 1=AP\left(5000-P\right)+B\left(P\right)=P\left(-A+B\right)+5000A\end{array}$

  $\begin{array}{l}\text{Comparing coefficient of }P\\ -A+B=0\\ \text{Comparing constant term}\end{array}$

  $\begin{array}{l}5000A=1\\ A=\frac{1}{5000}\\ \therefore B=A=\frac{1}{5000}\\ \therefore {\displaystyle \int \frac{1}{P\left(5000-P\right)}dP={\displaystyle \int {10}^{-5}dt}}\end{array}$

  $\begin{array}{l}{\displaystyle \int \frac{1/5000}{P}-\frac{1/5000}{P-5000}dP=}{\displaystyle \int {10}^{-5}dt}\\ \frac{1}{5000}\ln \left|P\right|-\frac{1}{5000}\ln \left|P-5000\right|={10}^{-5}t+C^{\prime }\end{array}$

  $\begin{array}{l}\frac{1}{5000}\ln \left|\frac{P}{P-5000}\right|={10}^{-5}t+C^{\prime }\\ \ln \left|\frac{P}{P-5000}\right|=0.05t+C^{″}\end{array}$

  $\begin{array}{l}\frac{P}{P-5000}=e^{0.05t+C^{″}}=e^{0.05t}{e}^{C^{″}}=C{e}^{0.05t}\\ \text{Now , it is given that }P=50\text{ when }t=0\end{array}$

  $\begin{array}{l}\therefore \frac{P}{P-5000}=C{e}^{0.05t}\\ \frac{50}{50-5000}=C{e}^{0.05\times 0}\\ C=-\frac{1}{99}\\ \therefore \frac{P}{P-5000}=-\frac{1}{99}{e}^{0.05t}\end{array}$

  $\begin{array}{l}P=-\frac{1}{99}{e}^{0.05t}\left(P-5000\right)=-\frac{1}{99}{e}^{0.05t}P+\frac{5000}{99}{e}^{0.05t}\\ P+\frac{1}{99}{e}^{0.05t}P=\frac{5000}{99}{e}^{0.05t}\end{array}$

  $\begin{array}{l}P\left(1+\frac{1}{99}{e}^{0.05t}\right)=\frac{5000}{99}{e}^{0.05t}\\ P\left(99+e^{0.05t}\right)=5000e^{0.05t}\\ P=\frac{5000e^{0.05t}}{99+e^{0.05t}}=\frac{5000}{99e^{-0.05t}+1}\end{array}$

Now, graphing the slope field of $\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)$

  

From the graph of slope field it conforms that $P=\frac{5000}{99e^{-0.05t}+1}$ is the solution of differential equation $\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)$ because the slope of $P$ matches with the slope field of given differential equation.