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Expert-verified Found in: Page 281 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Logistic Model. In Section 3.2 we discussed the logistic equation$$\frac{{{\bf{dp}}}}{{{\bf{dt}}}}{\bf{ = A}}{{\bf{p}}_{\bf{1}}}{\bf{p - A}}{{\bf{p}}^{\bf{2}}}{\bf{,p(0) = }}{{\bf{p}}_{\bf{o}}}$$and its use in modeling population growth. A more general model might involve the equation$$\frac{{{\bf{dp}}}}{{{\bf{dt}}}}{\bf{ = A}}{{\bf{p}}_{\bf{1}}}{\bf{p - A}}{{\bf{p}}^{\bf{r}}}{\bf{,p(0) = }}{{\bf{p}}_{\bf{o}}}$$where r>1. To see the effect of changing the parameter r in (25), take $${{\bf{p}}_{\bf{1}}}$$= 3, A = 1, and $${{\bf{p}}_{\bf{o}}}$$= 1. Then use a numerical scheme such as Runge–Kutta with h = 0.25 to approximate the solution to (25) on the interval$$0 \le {\bf{t}} \le 5$$ for r = 1.5, 2, and 3What is the limiting population in each case? For r >1, determine a general formula for the limiting population.

The general formula for the limiting population is$${{\bf{3}}^{\frac{{\bf{1}}}{{{\bf{r - 1}}}}}}$$.

See the step by step solution

## Step 1: Find the value of p

Here$${p_1} = 3$$, $$A = 1$$, and $${p_o} = 1$$ then the equation is:

$$\frac{{dp}}{{dt}} = 3p - {p^r}$$

Suppose$$r > 1$$. So,

$$\begin{array}{c}\int {\frac{1}{{3p - {p^r}}}dp} = \int {dt} \\\smallint \frac{1}{{{p^r}\left( {3{p^{1 - r}} - 1} \right)}}dp = t + c\\\int {\frac{1}{{3\left( {1 - r} \right)u}}du} = t + c\\\frac{{\ln \left| u \right|}}{{3\left( {1 - r} \right)}} = t + c\\\frac{{\ln \left| {3{p^{1 - r}} - 1} \right|}}{{3(1 - r)}} = t + c\\\ln \left| {3{p^{1 - r}} - 1} \right| = 3\left( {1 - r} \right)t + c\\{p^{1 - r}} = \frac{{C{e^{3(1 - r)t}} + 1}}{3}\\p = {\left( {\frac{{C{e^{3(1 - r)t}} + 1}}{3}} \right)^{\frac{1}{{1 - r}}}}\end{array}$$

Here $$p\left( 0 \right) = {p_o} = {\rm{ }}1$$ then

$$\begin{array}{c}{\bf{1 = }}{\left( {\frac{{{\bf{C + 1}}}}{{\bf{3}}}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\\{\bf{C = 2}}\\{\bf{p = }}{\left( {\frac{{{\bf{2}}{{\bf{e}}^{{\bf{3(1 - r)t}}}}{\bf{ + 1}}}}{{\bf{3}}}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\end{array}$$

## Step 2: Apply limits

$$\begin{array}{c}\mathop {{\bf{lim}}}\limits_{t \to \infty } {\bf{p(t) = }}\mathop {{\bf{lim}}}\limits_{t \to \infty } {\left( {\frac{{{\bf{2}}{{\bf{e}}^{{\bf{3(1 - r)t}}}}{\bf{ + 1}}}}{{\bf{3}}}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\\{\bf{ = }}{\left( {\frac{{2\mathop {{\bf{lim}}}\limits_{t \to \infty } {{\bf{e}}^{{\bf{3(1 - r)t}}}}{\bf{ + }}1}}{3}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\\{\bf{ = }}{\left( {\frac{{\bf{1}}}{{\bf{3}}}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\\{\bf{ = }}{{\bf{3}}^{\frac{{\bf{1}}}{{{\bf{r - 1}}}}}}\end{array}$$

## Step 3: Find the result by applying the Runge-Kutta method.

Apply this method in Mat Lab.

 T R=1.5 R=2 R=3 0 1 1 1 1 4.3024 2.7277 1.7254 1.5 6.1387 2.9343 1.7319 2 7.444 2.985 1.732 3 8.6129 2.999 1.7321 4 8.911 3 1.732 5 8.980 3 1.732

Therefore, the general formula for the limiting population is$${{\bf{3}}^{\frac{{\bf{1}}}{{{\bf{r - 1}}}}}}$$.

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