Chapter 6.5, Problem 45E

(a)

To determine

To find: $\frac{dP}{dt}$ is positive for $m<P<M$ and negative for $P<m$ or $P>M$ .

Explanation of Solution

Given information:

The extended logistic differential equation is given as:

  $\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\left(1-\frac{m}{P}\right)$

with $k>0$ the proportionality constant and $M$ the population maximum.

It is given that $m<P<M$

Calculation:

Rewrite the given condition as:

  $\begin{array}{l}m<P<M\\ \frac{m}{M}<\frac{P}{M}<1\end{array}$

Or

  $\frac{m}{P}<1<\frac{M}{P}$

Since $\frac{P}{M}<1$ then the differential equation will be positive as $k>0$ .

If $P<m$ then $1<\frac{m}{P}$ .

Which gives the second term of the differential equation $\left(1-\frac{m}{P}\right)$ negative value which gives the negative value of differential equation. Similarly, if $P>M$ then $\frac{P}{M}>1$ which makes the first term of differential equation negative so whole term will become negative.

(b)

To determine

To find: If $m=100$ , $M=1200$ , and assume that $m<P<M$ . Show that the differential equation can be rewritten 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:

The extended logistic differential equation is given as:

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

Calculation:

Substitute the value $m=100$ , $M=1200$ in the given differential equation gives:

  $\begin{array}{l}\frac{dP}{dt}=\frac{k}{M}\left(M-P\right)\left(P-m\right)\\ \frac{dP}{dt}=\frac{k}{1200}\left(1200-P\right)\left(P-100\right)\\ 1200\frac{dP}{dt}=k\left(1200-P\right)\left(P-100\right)\\ \frac{1}{\left(1200-P\right)\left(P-100\right)}\frac{dP}{dt}=\frac{k}{1200}\end{array}$

Solve the partial fraction $\frac{1}{\left(1200-P\right)\left(P-100\right)}$ :

  $\begin{array}{l}\frac{1}{\left(1200-P\right)\left(P-100\right)}=\frac{A}{\left(1200-P\right)}+\frac{B}{\left(P-100\right)}\\ 1=A\left(P-100\right)+B\left(1200-P\right)\end{array}$

Substitute $P=100$ gives:

  $\begin{array}{l}1=A\left(100-100\right)+B\left(1200-100\right)\\ 1=1100B\\ B=\frac{1}{1100}\end{array}$

Now substitute $P=1200$ gives:

  $\begin{array}{l}1=A\left(1200-100\right)+B\left(1200-1200\right)\\ 1=1100A\\ A=\frac{1}{1100}\end{array}$

Substitute the value of $A,B$ in $\frac{1}{\left(1200-P\right)\left(P-100\right)}=\frac{A}{\left(1200-P\right)}+\frac{B}{\left(P-100\right)}$

  $\frac{1}{\left(1200-P\right)\left(P-100\right)}=\frac{1}{1100\left(1200-P\right)}+\frac{1}{1100\left(P-100\right)}$

  $\frac{1}{\left(1200-P\right)\left(P-100\right)}\frac{dP}{dt}=\frac{k}{1200}$ can be written as:

  $\begin{array}{l}\left[\frac{1}{1100\left(1200-P\right)}+\frac{1}{1100\left(P-100\right)}\right]\frac{dP}{dt}=\frac{k}{1200}\\ \left[\frac{1}{\left(1200-P\right)}+\frac{1}{\left(P-100\right)}\right]\frac{1}{1100}\frac{dP}{dt}=\frac{k}{1200}\\ \left[\frac{1}{\left(1200-P\right)}+\frac{1}{\left(P-100\right)}\right]\frac{dP}{dt}=\frac{1100k}{1200}\\ \left[\frac{1}{\left(1200-P\right)}+\frac{1}{\left(P-100\right)}\right]\frac{dP}{dt}=\frac{11k}{12}\end{array}$

This is the required result.

(c)

To determine

To find: The solution of differential equation in part (b) when $P(0)=300$

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

Answer to Problem 45E

The solution of the differential equation in part ( C) is $\ln \left(\frac{P-100}{1200-P}\right)=\frac{11k}{12}t+\ln \left(\frac{2}{9}\right)$ .

Explanation of Solution

Given information:

From part (b) the differential equation is found $\left[\frac{1}{1200-P}+\frac{1}{P-100}\right]\frac{dP}{dt}=\frac{11}{12}k$

Calculation:

In order to solve the differential equation, it can be rewritten as:

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

Integrate both sides using the formula ${\displaystyle \int \frac{1}{x}=\ln x+c}$ .

  $\begin{array}{l}\left[{\displaystyle \int \frac{1}{\left(1200-P\right)}dp}+{\displaystyle \int \frac{1}{\left(P-100\right)}dp}\right]={\displaystyle \int \frac{11k}{12}dt}\\ -\ln \left(1200-P\right)+\ln \left(P-100\right)=\frac{11k}{12}t+c.\\ \ln \left(\frac{P-100}{1200-P}\right)=\frac{11k}{12}t+c\mathrm{.....}(1)\end{array}$

Substitute $P=300,t=0$ in equation $\left(1\right)$ :

  $\begin{array}{l}\ln \left(\frac{300-100}{1200-300}\right)=\frac{11k}{12}\cdot \left(0\right)+c.\\ c=\ln \left(\frac{200}{900}\right)\\ c=\ln \left(\frac{2}{9}\right)\end{array}$

Substitute the value of $c$ in equation $\left(1\right)$ :

  $\ln \left(\frac{P-100}{1200-P}\right)=\frac{11k}{12}t+\ln \left(\frac{2}{9}\right)$

The solution of the differential equation in part (c ) is $\ln \left(\frac{P-100}{1200-P}\right)=\frac{11k}{12}t+\ln \left(\frac{2}{9}\right)$ .

(d)

To determine

To find: Superimpose of the graph of the solution in part c with $k=0.1$ the slope field.

Answer to Problem 45E

The graph is given.

Explanation of Solution

Given information:

From part (b) the differential equation is found $\left[\frac{1}{1200-P}+\frac{1}{P-100}\right]\frac{dP}{dt}=\frac{11}{12}k$

The solution of this differential equation is $\ln \left(\frac{P-100}{1200-P}\right)=\frac{11k}{12}t+\ln \left(\frac{2}{9}\right)$ .

Substitute $k=0.1$ in the solution gives:

  $\begin{array}{l}\ln \left(\frac{P-100}{1200-P}\right)=\frac{11\left(0.1\right)}{12}t+\ln \left(\frac{2}{9}\right)\\ \ln \left(\frac{P-100}{1200-P}\right)=\frac{1.1}{12}t+\ln \left(\frac{2}{9}\right)\end{array}$

The slope field graph is given below:

  

The graph of the solution $\ln \left(\frac{P-100}{1200-P}\right)=\frac{1.1}{12}t+\ln \left(\frac{2}{9}\right)$ is given below:

  

(e)

To determine

To find: The general extended equation with restriction $m<P<M$ .

Answer to Problem 45E

The solution of the differential equation $-\ln (M-P)+\ln (P-m)=\left(M-m\right)\frac{k}{M}t+c$ .

Explanation of Solution

Given information:

The general differential equation is given as:

  $\frac{dP}{dt}=\frac{k}{M}\left(M-P\right)\left(P-m\right)$

This equation can be rewritten as:

  $\frac{dP}{\left(M-P\right)\left(P-m\right)}=\frac{k}{M}dt$

Solve the fractional term by partial fraction method.

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

Substitute $M=P$ gives:

  $\begin{array}{l}1=A\left(M-m\right)\\ A=\frac{1}{\left(M-m\right)}\end{array}$

Substitute $m=P$ gives:

  $\begin{array}{l}1=B\left(M-m\right)\\ B=\frac{1}{\left(M-m\right)}\end{array}$

Substitute these values in $\frac{1}{\left(M-P\right)\left(P-m\right)}=\frac{A}{\left(M-P\right)}+\frac{B}{\left(P-m\right)}$ gives:

  $\frac{1}{\left(M-P\right)\left(P-m\right)}=\frac{1}{\left(M-m\right)}\left[\frac{1}{\left(M-P\right)}+\frac{1}{\left(P-m\right)}\right]$

Now the differential equation can be written as:

  $\frac{1}{\left(M-m\right)}\left[\frac{1}{\left(M-P\right)}+\frac{1}{\left(P-m\right)}\right]dp=\frac{k}{M}dt$

Integrate on both sides:

  $\begin{array}{l}\left[{\displaystyle \int \frac{dP}{\left(M-P\right)}}+{\displaystyle \int \frac{dP}{\left(P-m\right)}}\right]=\left(M-m\right)\frac{k}{M}dt\\ -\ln (M-P)+\ln (P-m)=\left(M-m\right)\frac{k}{M}t+c\end{array}$

Apply the formula ${\displaystyle \int \frac{1}{(1-x)}dx=-\ln (1-x)+c}$

This is the general solution of the given equation.