Q7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Consider the differential equation  $\frac{\mathbf{dp}}{\mathbf{dt}}\mathbf{=}\mathbf{p}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{p}\mathbf{)}$  for the population p (in thousands) of a certain species at time t.

⦁    Sketch the direction field by using either a computer software package or the method of isoclines.

⦁    If the initial population is 4000 [that is, $p\left(\mathbf{0}\right)=\mathbf{4}$], what can you say about the limiting population  ${\mathbf{lim}}_{\mathbf{t}\to \infty }\mathbf{P}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{?}$

⦁    If $\mathit{p}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{7}$ , what is  ${\mathbf{lim}}_{\mathbf{t}\to \infty }\mathbf{P}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{?}$

⦁    If $\mathit{p}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}$ , what is  ${\mathbf{lim}}_{\mathbf{t}\to \infty }\mathbf{P}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{?}$

⦁    Can a population of 900 ever increase to 1100?

Step-by-Step Solution

Answer

⦁    The Sketch is drawn for the direction field

⦁    2

⦁    2

⦁    0

⦁    No

1(a): Drawing the Sketch for the direction field

Hence the sketch is drawn for the direction filed.

Solving the differential equation

$$\begin{gathered} \frac{\mathbf{dp}}{\mathbf{p}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{p}\mathbf{)}}\mathbf{=}\mathbf{dt} \\ \frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}\int \frac{\mathbf{dp}}{\mathbf{p}}\mathbf{+}\int \frac{\mathbf{dp}}{\mathbf{p}\mathbf{-}\mathbf{1}}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}\int \frac{\mathbf{dp}}{\mathbf{p}\mathbf{-}\mathbf{2}}\mathbf{=}\int \mathbf{dt} \\ \mathbf{log}\left|\mathbf{p}\right|\mathbf{+}\mathbf{log}{\left|\mathbf{p}\mathbf{-}\mathbf{1}\right|}^{\mathbf{-}\mathbf{2}}\mathbf{+}\mathbf{log}{\left|\mathbf{p}\mathbf{-}\mathbf{2}\right|}^{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{c} \\ \mathbf{log}\left|\frac{\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\right|\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{c} \\ \frac{\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}} \end{gathered}$$

3(b): Putting the initial condition p 0 = 4 in the solution

$$\begin{gathered} \frac{\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{(}\mathbf{4}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{4}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{.}\mathbf{1} \\ \frac{\mathbf{32}}{\mathbf{9}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}} \\ \frac{\mathbf{p}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{32}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}} \\ {\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\frac{\mathbf{p}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\frac{\mathbf{32}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}} \\ {\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2} \end{gathered}$$

Hence the limiting population is 2.

4(c): Placing the first state p ( 0 ) = 1 . 7 in the solution found in Step 2

$$\begin{gathered} \frac{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\left)}\mathbf{\right(}\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}·\mathbf{1} \\ \frac{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{0459}}{\mathbf{0}\mathbf{.}\mathbf{49}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}} \\ \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{094}\mathbf{=}{\mathbf{c}}_{\mathbf{1}} \\ \frac{\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{094}\mathbf{)}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}} \\ {\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\frac{\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{094}\mathbf{)}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}} \\ {\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2} \end{gathered}$$

$\mathit{H}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{l}\mathit{i}\mathit{m}\mathit{i}\mathit{t}\mathit{i}\mathit{n}\mathit{g}\mathbf{ }\mathit{p}\mathit{o}\mathit{p}\mathit{u}\mathit{l}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{i}\mathit{s}\mathbf{ }\mathbf{2}\mathbf{.}$

5(d): Positioning the primary order in the solution found in Step 2

$$\begin{gathered} \frac{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}·\mathbf{1} \\ \frac{\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{3824}}{\mathbf{0}\mathbf{.}\mathbf{04}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}} \\ \mathbf{-}\mathbf{34}\mathbf{.}\mathbf{56}\mathbf{=}{\mathbf{c}}_{\mathbf{1}} \\ \frac{\mathbf{p}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{34}\mathbf{.}\mathbf{56}\mathbf{)}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}} \\ {\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\frac{\mathbf{p}\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\mathbf{(}\mathbf{-}\mathbf{34}\mathbf{.}\mathbf{56}\mathbf{)}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}} \\ {\mathbf{lim}}_{\mathbf{t}\to \mathbf{+}\infty }\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

$\mathit{H}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{l}\mathit{i}\mathit{m}\mathit{i}\mathit{t}\mathit{i}\mathit{n}\mathit{g}\mathbf{ }\mathit{p}\mathit{o}\mathit{p}\mathit{u}\mathit{l}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{i}\mathit{s}\mathbf{ }\mathbf{0}\mathbf{.}$

6(e): Analyzing the direction field graph

As it can be seen in the graph given in part (a), that the points 0,1, 2 are terminal points, and every point less than 1 cannot increase to a point greater than 1.

Now, 900 is equivalent to 0.9 (in thousands) while 1100 is equivalent to 1.1 (in thousands).

Hence a population of 900 can never increase to 1100.