Chapter 7, Problem 61RE

To determine

To Find:

To find the exact solution of $\frac{dP}{dt}=0.002P\left(1-\frac{P}{800}\right)$ with the initial conditions $P\left(0\right)=50$ using analytic method.

Answer to Problem 61RE

The exact solution is $P=\frac{800}{1+15e^{-0.002t}}$ .

Explanation of Solution

Given information:

The differential equation is $\frac{dP}{dt}=0.002P\left(1-\frac{P}{800}\right)$ with initial conditions $P\left(0\right)=50$ .

The logistic differential equation is

  $\begin{array}{l}\frac{dP}{dt}=kP\left(M-P\right)\\ \frac{dP}{P\left(M-P\right)}=kdt\end{array}$

Integrating with respect to t,

  $\begin{array}{l}{\displaystyle \int \frac{dP}{P\left(M-P\right)}}={\displaystyle \int kdt}\\ {\displaystyle \int \frac{dP}{P\left(M-P\right)}}=kt+C\\ {\displaystyle \int \left(\frac{1}{MP}+\frac{1}{M\left(M-P\right)}\right)dP}=kt+A\\ \frac{1}{M}\ln P-\frac{1}{M}\ln \left|P-M\right|=kt+A\\ \ln \left|\frac{P}{M-P}\right|=Mkt+A\\ \frac{P}{M-P}=A{e}^{Mkt}\\ P=MA{e}^{Mkt}-PA{e}^{Mkt}\\ P\left(1+A{e}^{Mkt}\right)=MA{e}^{Mkt}\\ P=\frac{MA{e}^{Mkt}}{\left(1+A{e}^{Mkt}\right)}\\ P=\frac{M}{\left(1+A{e}^{-Mkt}\right)}\\ \frac{dP}{dt}=2.5\times {10}^{-6}P\left(800-P\right)\\ k=2.5\times {10}^{-6};M=800\\ P=\frac{800}{1+A{e}^{-0.002t}}\\ 50=\frac{800}{1+A}\\ \Rightarrow A=15\\ \Rightarrow P=\frac{800}{1+15e^{-0.002t}}\end{array}$

Therefore, the exact solution is $P=\frac{800}{1+15e^{-0.002t}}$ .