Chapter 7.4, Problem 21E

a.

To determine

To find : the solution of the equation that satisfies the initial condition $P\left(0\right)=P_0$

Answer to Problem 21E

The solution of the equation that satisfies the initial condition $P\left(0\right)=P_0$ is $P\left(t\right)=\left(P_0-\frac{m}{k}\right)e^{kt}+\frac{m}{k}$

Explanation of Solution

Given information : The question is

The differential equation $\frac{dP}{dt}=kP-m\text{ ;where }k=\alpha -\beta$

Calculation : the solution of the equation

  $\begin{array}{l}\frac{dP}{dt}=kP-m\\ \frac{dP}{kP-m}=dt\\ {\displaystyle \int \frac{dP}{kP-m}={\displaystyle \int dt}}\\ \frac{1}{k}{\displaystyle \int \frac{kdP}{kP-m}}={\displaystyle \int dt}\\ \frac{1}{k}{\displaystyle \int \frac{dx}{x}}={\displaystyle \int dt}\\ \frac{1}{k}\ln \left|x\right|=t+C\\ \frac{1}{k}\ln \left|kP-m\right|=t+C\left\{t=0;P=P_0\right\}\\ \frac{1}{k}\ln \left|k{P}_0-m\right|=C\\ \frac{1}{k}\ln \left|kP-m\right|=t+\frac{1}{k}\ln \left|k{P}_0-m\right|\\ \ln \left|kP-m\right|=kt+\ln \left|k{P}_0-m\right|\\ \ln \left|kP-m\right|-\ln \left|k{P}_0-m\right|=kt\\ \ln \frac{\left|kP-m\right|}{\left|k{P}_0-m\right|}=kt\\ \frac{\left|kP-m\right|}{\left|k{P}_0-m\right|}=e^{kt}\\ kP-m=\left(k{P}_0-m\right)e^{kt}\\ kP=\left(k{P}_0-m\right)e^{kt}+m\\ P\left(t\right)=\left(P_0-\frac{m}{k}\right)e^{kt}+\frac{m}{k}\end{array}$

b.

To determine

To calculate : the condition on $m$ will lead to an exponential of the population

Answer to Problem 21E

The population will grow if $m<k{P}_0$

Explanation of Solution

Given information : The question is

The differential equation $\frac{dP}{dt}=kP-m\text{ ;where }k=\alpha -\beta$

Calculation : from part (a)

  $\begin{array}{l}P\left(t\right)=\left(P_0-\frac{m}{k}\right)e^{kt}+\frac{m}{k}\\ \frac{dP}{dt}=kP-m\\ \frac{dP}{dt}=k\left[\left(P_0-\frac{m}{k}\right)e^{kt}+\frac{m}{k}\right]-m\\ \frac{dP}{dt}=\left[\left(k{P}_0-m\right)e^{kt}+m\right]-m\\ \frac{dP}{dt}=\left(k{P}_0-m\right)e^{kt}\end{array}$

When $\frac{dP}{dt}$ is positive, then $P$ increases as $t$ decreases

When $\frac{dP}{dt}$ is negative, then $P$ decreases as $t$ increase

Therefore, population will grow exponentially if $\frac{dP}{dt}>0$

  $\begin{array}{l}\frac{dP}{dt}>0\Rightarrow \left(k{P}_0-m\right)e^{kt}>0\\ k{P}_0-m>0\\ k{P}_0>0\\ m<k{P}_0\end{array}$

c.

To determine

To calculate : the condition on $m$ will result in a constant population or a decline population

Answer to Problem 21E

Population will remain constant if $m=k{P}_0$

Population will decline if $m>k{P}_0$

Explanation of Solution

Given information : The question is

The differential equation $\frac{dP}{dt}=kP-m\text{ ;where }k=\alpha -\beta$

Calculation : from part (a)

  $\begin{array}{l}P\left(t\right)=\left(P_0-\frac{m}{k}\right)e^{kt}+\frac{m}{k}\\ \frac{dP}{dt}=kP-m\\ \frac{dP}{dt}=k\left[\left(P_0-\frac{m}{k}\right)e^{kt}+\frac{m}{k}\right]-m\\ \frac{dP}{dt}=\left[\left(k{P}_0-m\right)e^{kt}+m\right]-m\\ \frac{dP}{dt}=\left(k{P}_0-m\right)e^{kt}\end{array}$

When $\frac{dP}{dt}$ is positive, then $P$ increases as $t$ decreases

When $\frac{dP}{dt}$ is negative, then $P$ decreases as $t$ increase

When $\frac{dP}{dt}$ is zero, then $P$ remain constant as $t$ increase

Population will remain constant if $\frac{dP}{dt}=0$

  $\begin{array}{l}\frac{dP}{dt}=0\Rightarrow \left(k{P}_0-m\right)e^{kt}=0\\ k{P}_0-m=0\\ m=k{P}_0\end{array}$

Population will decline if $\frac{dP}{dt}<0$

  $\begin{array}{l}\frac{dP}{dt}<0\Rightarrow \left(k{P}_0-m\right)e^{kt}<0\\ k{P}_0-m<0\\ m>k{P}_0\end{array}$

d.

To determine

To calculate : the population expanding or declining at the time

Answer to Problem 21E

The population is declining

Explanation of Solution

Given information : The question is

In 1847, population of Ireland was $8$ million

Formula used : $k=\frac{\text{Difference between birth and death rates}}{\text{Population}}$

Calculation : It is given that, difference between the relative birth and death rates was $1.6\%$ of the population

Therefore,

  $k=0.016$

  $210,000$ inhabitants per emigrated from Ireland

Therefore,

  $\begin{array}{l}m=0.21\times {10}^6\\ k{P}_0=0.016\times \left(8\times {10}^6\right)\\ k{P}_0=0.128\times {10}^6\end{array}$