Chapter 7.5, Problem 45E

a.

To determine

Show that dP/dt is positive for M < P < M and negative if P < M or P > M.

Explanation of Solution

Given information: Condition leads to the extended logistic differential equation.

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\left(1-\frac{m}{p}\right)\\ =\frac{k}{M}\left(M-P\right)\left(p-m\right)\end{array}$

Proof:

For m < P < M, both terms of (M − P)(P − m) are positive, and thus dp/dt is positive overall. However, if P > M or m < P one of these terms is negative, and thus the derivative is negative.

b.

To determine

Show that the differential equation is written in the form.

  $\left[\frac{1}{1200-P}+\frac{1}{P-100}\right]\frac{dP}{dt}=\frac{11}{12}k$

Explanation of Solution

Given information: Condition leads to the extended logistic differential equation.

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\left(1-\frac{m}{p}\right)\\ =\frac{k}{M}\left(M-P\right)\left(p-m\right)\end{array}$

Proof:

Using the Heaviside coverup method when appropriate,

  $\begin{array}{l}\frac{dP}{dt}=\frac{k}{M}\left(M-P\right)\left(P-m\right)\\ \frac{dP}{\left(M-P\right)\left(P-m\right)}=\frac{k}{M}dt\\ \left(\frac{1}{\left(P-m\right)\left(M-P\right)}-\frac{1}{\left(P-m\right)\left(M-P\right)}\right)dP=\frac{k}{M}dt\\ \left(\frac{1}{\left(P-100\right)\left(1200-100\right)}-\frac{1}{\left(1200-P\right)\left(1200-100\right)}\right)dP=\frac{k}{1200}dt\\ \left(\frac{1}{P-100}+\frac{1}{1200-P}\right)dP=\frac{11}{12}dt\\ \ln \left|P-100\right|-\ln \left|1200-P\right|=\frac{11}{12}kt+C\\ \ln \left|\left(\frac{P-100}{1200-P}\right)\right|=\frac{11}{12}kt+C\\ \ln \left|\left(\frac{1100}{1200-P}-1\right)\right|=\frac{11}{12}kt+C\\ \frac{1100}{1200-P}-1=C{e}^{\frac{11}{12}kt}\\ 1200-P=\frac{1100}{1+C{e}^{\frac{11}{12}kt}}\\ P=\frac{100+1200C{e}^{\frac{11}{12}kt}}{1+C{e}^{\frac{11}{12}kt}}\end{array}$

c.

To determine

To find the solution to part (b) that satisfies P(0) = 300.

Answer to Problem 45E

The required solution is, $P=\frac{100+\frac{800}{3}C{e}^{\frac{11}{12}kt}}{1+\frac{800}{3}{e}^{\frac{11}{12}kt}}$

Explanation of Solution

Given information: Condition leads to the extended logistic differential equation.

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\left(1-\frac{m}{p}\right)\\ =\frac{k}{M}\left(M-P\right)\left(p-m\right)\end{array}$

Calculation: The required solution is obtained as,

  $\begin{array}{l}P=\frac{100+1200C{e}^{\frac{11}{12}kt}}{1+C{e}^{\frac{11}{12}kt}}\\ 300=\frac{100+1200C}{1+C}\\ 300+300C=100+1200C\\ C=2/9\end{array}$

Substituting $C=2/9$ in equation $P=\frac{100+1200C{e}^{\frac{11}{12}kt}}{1+C{e}^{\frac{11}{12}kt}}$

  $P=\frac{100+\frac{800}{3}C{e}^{\frac{11}{12}kt}}{1+\frac{800}{3}{e}^{\frac{11}{12}kt}}$

Hence, the required solution is $P=\frac{100+\frac{800}{3}C{e}^{\frac{11}{12}kt}}{1+\frac{800}{3}{e}^{\frac{11}{12}kt}}$ .

d.

To determine

Superimpose the graph of the solution in part (C) with k = 0.1 on a slope field of the differential equation.

Explanation of Solution

Given information: Condition leads to the extended logistic differential equation.

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\left(1-\frac{m}{p}\right)\\ =\frac{k}{M}\left(M-P\right)\left(p-m\right)\end{array}$

Graph: The graph is,

  

e.

To determine

To solve the general extended differential equation with the restriction m < P < M.

Answer to Problem 45E

The required solution is $P=\frac{m+MC{e}^{\frac{M-m}{M}kt}}{1+C{e}^{\frac{M-m}{M}kt}}$ .

Explanation of Solution

Given information: Condition leads to the extended logistic differential equation.

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\left(1-\frac{m}{p}\right)\\ =\frac{k}{M}\left(M-P\right)\left(p-m\right)\end{array}$

Calculation: The required solution is obtained as,

  $\begin{array}{l}\frac{dP}{dt}=\frac{k}{M}\left(M-P\right)\left(P-m\right)\\ \frac{dP}{\left(M-P\right)\left(P-m\right)}=\frac{k}{M}dt\\ \left(\frac{1}{\left(P-m\right)\left(M-m\right)}-\frac{1}{\left(P-m\right)\left(M-m\right)}\right)dP=\frac{k}{M}dt\\ \left(\frac{1}{\left(P-m\right)\left(M-m\right)}+\frac{1}{\left(M-P\right)\left(M-m\right)}\right)dP=\frac{k}{M}dt\\ \left(\frac{1}{\left(P-m\right)}+\frac{1}{\left(M-P\right)}\right)dP=\frac{M-m}{M}k\\ \ln \left|P-m\right|-\ln \left|1200-P\right|=\frac{M-m}{M}kt+C\\ \ln \left|\left(\frac{P-100}{M-P}\right)\right|=\frac{M-m}{M}kt+C\\ \ln \left|\left(\frac{M-m}{M-P}-1\right)\right|=\frac{M-m}{M}kt+C\\ \frac{M-m}{M-P}-1=C{e}^{\frac{M-m}{M}kt}\\ M-P=\frac{M-m}{1+C{e}^{\frac{M-m}{M}kt}}\\ P=\frac{m+MC{e}^{\frac{M-m}{M}kt}}{1+C{e}^{\frac{M-m}{M}kt}}\end{array}$

Hence, the required solution is $P=\frac{m+MC{e}^{\frac{M-m}{M}kt}}{1+C{e}^{\frac{M-m}{M}kt}}$ .