/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 The system of differential equat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 3

The system of differential equations $$ \begin{array}{l}{\frac{d x}{d t}=0.5 x-0.004 x^{2}-0.001 x y} \\ {\frac{d y}{d t}=0.4 y-0.001 y^{2}-0.002 x y}\end{array} $$ $$ \begin{array}{l}{\text { is a model for the populations of two species. }} \\\ {\text { (a) Does the model describe cooperation, or competition, }} \\\ {\text { or a predator-prey relationship? }} \\ {\text { (b) Find the equilibrium solutions and explain their }} \\ {\text { significance. }}\end{array} $$

Short Answer

Expert verified
(a) The model describes competition. (b) Equilibrium solutions are \((0, 0)\) and \((125, 150)\) indicating extinction and coexistence, respectively.

Step by step solution

01

- Understanding the Model

The given system of differential equations represents the interaction between two species based on their population sizes, denoted by \(x\) and \(y\). Each equation describes the rate of change of each species' population over time, involving interactions between the species, represented by terms containing both \(x\) and \(y\).
02

- Analyzing the Interaction Types

To determine the type of interaction (cooperation, competition, or predator-prey), examine the cross terms in the equations: \(-0.001xy\) and \(-0.002xy\). These terms decrease the growth rate of each species due to the presence of the other, suggesting a competitive relationship as both negatively impact each other's growth.
03

- Finding Equilibrium Solutions

Equilibrium solutions occur when the growth rates are zero, i.e., \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\). Set the given equations: \[0.5x - 0.004x^2 - 0.001xy = 0\] and \[0.4y - 0.001y^2 - 0.002xy = 0\] to find values of \(x\) and \(y\) at equilibrium.
04

- Solving for Equilibrium Points

Solving \(0.5x - 0.004x^2 - 0.001xy = 0\), we can factor out \(x\): \[x(0.5 - 0.004x - 0.001y) = 0\]. This gives equilibria at \(x = 0\) or \(0.5 - 0.004x - 0.001y = 0\). Similar procedure for \(y\): \[y(0.4 - 0.001y - 0.002x) = 0\] leading to \(y = 0\) or \(0.4 - 0.001y - 0.002x = 0\).
05

- Finding Equilibria

Setting equations from Step 4 to zero gives possible equilibria: \((x, y) = (0, 0)\), when both populations are extinct, and solving the two linear equations \(0.5 = 0.004x + 0.001y\) and \(0.4 = 0.002x + 0.001y\) for other equilibrium points.
06

- Solving the Linear System

Using linear algebra or substitution, solve: \[x = 125, y = 150\] from equations \(0.5 = 0.004x + 0.001y\) and \(0.4 = 0.002x + 0.001y\), providing another equilibrium point where both populations coexist.
07

- Significance of Equilibria

The equilibria \((0, 0)\) indicates extinction for both species. The equilibrium \((125, 150)\) suggests a stable population size where both species can coexist, despite competitive interaction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Dynamics
Population dynamics involves the study of how populations of species change over time and space. It is a key aspect in the study of ecology as it looks at how populations interact with each other and their environments. For instance, in a system of differential equations, such as - \( \frac{d x}{d t} = 0.5 x - 0.004 x^{2} - 0.001 x y \)- \( \frac{d y}{d t} = 0.4 y - 0.001 y^{2} - 0.002 x y \),these equations represent the growth rates of two species.In this model:
  • \(x\) and \(y\) are the population sizes of two different species.
  • Each term in the equations influences how these populations grow over time.
  • The coefficients (like 0.5 or 0.004) determine the influence of each factor, such as natural growth, competition, or other interactions.
Understanding these interactions helps ecologists predict future population sizes and make informed decisions on conservation and resource management.
Equilibrium Solutions
Equilibrium solutions in differential equations occur when the rates of change in populations are zero. This means that the populations reach a state where they remain constant. For the given model, equilibrium points are found by setting:- \( \frac{dx}{dt} = 0 \)- \( \frac{dy}{dt} = 0 \).Upon solving, we find solutions such as:
  • \((0,0)\) - Both populations are extinct.
  • \((125,150)\) - Populations coexist in a stable state.
These solutions are significant because they indicate potential long-term behavior of populations under specific conditions. Analyzing equilibria helps us understand whether a system might reach a balanced state or face risks like extinction.
Predator-Prey Model
While this system involves competition, it contrasts with the classic predator-prey model. In predator-prey models, one species (the predator) benefits by preying on another species (the prey), resulting in different interaction terms such as
  • a positive impact on the predator's growth rate from the prey presence.
  • a negative impact on the prey's growth due to predation.
In predator-prey systems, equations typically have form: - prey growth reduced by predator interactions, - predator growth increased by prey capture. Analyzing such models helps understand cycles in ecosystems, where prey abundance affects predator populations and vice versa. Unlike competition, predator-prey interactions can stabilize populations through natural checks and balances.
Competition in Ecosystems
Competition occurs when two species negatively impact each other's growth, often because of shared resources. In our model, illustrated by the terms - \(-0.001xy\) and \(-0.002xy\),both species compete, resulting in the inhibition of growth for each other. This is common in nature where organisms vie for limited resources like food, space, or sunlight.Competitive interactions help us comprehend:
  • how species populations can fluctuate impacts community structures.
  • the dynamic balance between competing organisms which influences biodiversity.
Analyzing competition informs decisions on managing ecosystems sustainably, ensuring species do not drive each other to extinction due to over-competition.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a seasonal-growth model, a periodic function of time is introduced to account for seasonal variations in the rate of growth. Such variations could, for example, be caused by seasonal changes in the availability of food. (a) Find the solution of the seasonal-growth model $$ \frac{d P}{d t}=k P \cos (r t-\phi) \quad P(0)=P_{0} $$ where \(k, r,\) and \(\phi\) are positive constants. (b) By graphing the solution for several values of \(k, r,\) and \(\phi,\) explain how the values of \(k, r,\) and \(\phi\) affect the solution. What can you say about \(\lim _{t \rightarrow \infty} P(t) ?\)

Find the solution of the differential equation that satisfies the given initial condition. $$ x+3 y^{2} \sqrt{x^{2}+1} \frac{d y}{d x}=0, \quad y(0)=1 $$

One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction \(y\) of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by \(y .\) (b) Solve the differential equation. (c) A small town has 1000 inhabitants. At \(8 \mathrm{AM}, 80\) people have heard a rumor. By noon half the town has heard it. At what time will \(90 \%\) of the population have heard the rumor?

Solve the differential equation. $$ x y y^{\prime}=x^{2}+1 $$

(a) Assume that the carrying capacity for the US population is 800 million. Use it and the fact that the population was 282 million in 2000 to formulate a logistic model for the US population. (b) Determine the value of \(k\) in your model by using the fact that the population in 2010 was 309 million. (c) Use your model to predict the US population in the years 2100 and 2200 . (d) Use your model to predict the year in which the US population will exceed 500 million.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.