Chapter 7.6, Problem 9E

To determine

To sketch: The solution curve that passes through the point (1000, 40).

Answer to Problem 9E

The solutioncurve matches the curve obtained from the direction field.

Explanation of Solution

Given information:

Consider the following differential equation which models the rabbit and wolf population:

  $\frac{dW}{dR}=\frac{-0.02W+0.00002RW}{0.08R-0.001RW}$

Formula used:

The graph is plotted against x axis and y axis.

Calculation:

Where W and R are the population of which and rabbit respectively

This differential equation can be written as:

  $\begin{array}{l}\frac{dW}{dR}=\frac{-0.02W+0.0002RW}{0.08R-0.001RW}\\ =\frac{W}{R}\left(\frac{-0.02+0.00002R}{0.08-0.001W}\right)\\ \frac{dW\left(0.08-0.001W\right)}{W}=\frac{\left(-0.02+0.00002R\right)dR}{R}\\ \frac{\left(0.08-0.001W\right)}{W}dW=\frac{\left(-0.02+0.00002R\right)}{R}dR\end{array}$

Now integrate on both sides to get:

  $\begin{array}{l}{\displaystyle \int \frac{\left(0.08-0.001W\right)}{W}}dW={\displaystyle \int \frac{\left(-0.02+0.00002R\right)}{R}dR}\\ {\displaystyle \int \frac{\left(0.08\right)}{W}}dW-{\displaystyle \int \frac{0.001W}{W}}={\displaystyle \int \frac{\left(-0.02\right)}{R}}dR+{\displaystyle \int \frac{0.00002R}{R}}dR\\ 0.08\ln \left(W\right)-0.001W=-0.02\ln R+0.00002R+C\\ 0.08\ln \left(W\right)+0.02\ln R=0.00002R+0.001W+C\end{array}$

Using the property a In (x) = In x^a, one can write:

  $\begin{array}{l}0.08\ln \left(W\right)+0.02\ln R=0.00002R+0.001W+C\\ \ln W^{0.08}+\ln R^{0.02}=0.00002R+0.001W+C\\ \ln \left(W^{0.08}{R}^{0.02}\right)=0.00002R+0.001W+C\\ W^{0.08}{R}^{0.02}=e^{\left(0.00002R+0.001W+C\right)}\end{array}$

Then,

  $\begin{array}{l}{W}^{0.08}{R}^{0.02}=e^{\left(0.00002R+0.001W+C\right)}\\ W^{0.08}{R}^{0.02}=e^{\left(0.00002R\right)}{e}^{0.001W}{e}^C\\ \frac{W^{0.08}{R}^{0.02}}{e^{\left(0.00002R\right)}{e}^{0.001W}}=c\end{array}$

Where c is a constant $e^C$

When the curve of the implicit plot passes through (1000, 40) , the value of the constant c is:

  $\begin{array}{l}\frac{W^{0.08}{R}^{0.02}}{e^{\left(0.00002R\right)}{e}^{0.001\left(40\right)}}=c\\ c=1.45\end{array}$

So the equation of a curve passing through (1000,40)will be:

  $\frac{W^{0.08}{R}^{0.02}}{e^{\left(0.000012R\right)}{e}^{0.001W}}=1.45$

For the implicit plot in maple, use the following commands:

with (plots, implicitplot)

  $implicitplot\left(\frac{x^{0.02}{y}^{0.08}}{\exp \left(0.00002x\right)\exp \left(0.001y\right)}=1.45,x=\mathrm{0..4000},y=\mathrm{0..180}\right)$

Then the plot will be:

  

The solutioncurve matches the curve obtained from the direction field.

Conclusion:

The solutioncurve matches the curve obtained from the direction field.