Chapter 7, Problem 22RE

a.

To determine

that which variable represents bird population or insect population

Answer to Problem 22RE

  $x$ Represents insect population

  $y$ Represents bird population

Explanation of Solution

Given information : The question is

Population of birds and insects are modeled by equation

  $\begin{array}{l}\frac{dx}{dt}=0.4x-0.002xy\mathrm{......}\left(1\right)\\ \frac{dy}{dt}=-0.2y+0.000008xy\mathrm{......}\left(2\right)\end{array}$

Calculation :

In the absence of species $y$ , the population of $x$ is increasing

Absence of species $y$ means $y=0$

After substituting $y=0$ in equation $\left(1\right)$

  $\frac{dx}{dt}=0.4x$

Then $\frac{dx}{dt}$ is positive, therefore $x$ is increasing

In the absence of species $x$ , the population of $y$ is decreasing

Absence of species $x$ means $x=0$

After substituting $x=0$ in equation $\left(2\right)$

  $\frac{dy}{dt}=-0.2y$

Then $\frac{dy}{dt}$ is negative, therefore $x$ is decreasing

Therefore,

  $x$ Represents insect population

  $y$ Represents bird population

b.

To determine

To find : an equilibrium solution and explain their significance

Answer to Problem 22RE

The equilibrium solutions are

  $\begin{array}{l}x=0\&y=0\\ x=2500\&y=200\end{array}$

Explanation of Solution

Given information : The question is

Population of birds and insects are modeled by equation

  $\begin{array}{l}\frac{dx}{dt}=0.4x-0.002xy\mathrm{......}\left(1\right)\\ \frac{dy}{dt}=-0.2y+0.000008xy\mathrm{......}\left(2\right)\end{array}$

Calculation : The equilibrium solution is the set of values of $x\&y$ for which both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are zero

  $\begin{array}{l}\frac{dx}{dt}=0\\ 0.4x-0.002xy=0\\ x=0\\ 200-y=0\\ y=200\\ \frac{dx}{dt}=0,\text{when }y=200\text{ or }x=0\\ \frac{dy}{dt}=0\\ -0.2y+0.000008xy=0\\ y=0\\ -2500+x=0\\ x=2500\\ \frac{dy}{dt}=0,\text{when }y=0\text{ or }x=2500\\ x=0\&y=0\\ x=2500\&y=200\end{array}$

c.

To determine

To find : an expression for $\frac{dy}{dx}$

Answer to Problem 22RE

The expression $\frac{dy}{dx}=\frac{-0.2y+0.000008xy}{0.4x-0.002xy}$

Explanation of Solution

Given information : The question is

Population of birds and insects are modeled by equation

  $\begin{array}{l}\frac{dx}{dt}=0.4x-0.002xy\mathrm{......}\left(1\right)\\ \frac{dy}{dt}=-0.2y+0.000008xy\mathrm{......}\left(2\right)\end{array}$

Calculation :

From the chain rule

  $\begin{array}{l}\frac{dy}{dt}\times \frac{dt}{dx}=\frac{dy}{dx}\\ \frac{dx}{dt}=0.4x-0.002xy\\ \frac{dy}{dt}=-0.2y+0.000008xy\\ \therefore \frac{dy}{dx}=\left(-0.2y+0.000008xy\right)\times \frac{1}{0.4x-0.002xy}\\ \frac{dy}{dx}=\frac{-0.2y+0.000008xy}{0.4x-0.002xy}\end{array}$

d.

To determine

To use : the phase trajectory to describe how both population change

Explanation of Solution

Given information : The question is

Population of birds and insects are modeled by equation

  $\begin{array}{l}\frac{dx}{dt}=0.4x-0.002xy\mathrm{......}\left(1\right)\\ \frac{dy}{dt}=-0.2y+0.000008xy\mathrm{......}\left(2\right)\end{array}$

The phase trajectory, corresponding to the initial condition

To draw the phase trajectory use direction field

  

The arrow indicates that as $t$ increases the point representing the two population on the phase plot will move in the counter clockwise direction