Explanation Given: The condition for x and y are x m < x < x M and y 1 < a b < y 2 . Calculation: The plane autonomous system for Lotka-Volterra model is as follows: x ′ = x ( − a + b y ) y ′ = y ( − c x + d ) In the above equations x denotes the number of predators and y denotes the number of prey. As per the phase plane method the linear differential equation is obtained as follows: d y d x = d y d t d t d x = y ′ x ′ Substitute the expression for x ′ and y ′ in the above equation. d y d x = y ′ x ′ = y ( − c x + d ) x ( − a + b y ) Separate the variables in the above equation and integrate as follows: d y d x = y ( − c x + d ) x ( − a + b y ) ( − a + b y ) d y y = ( − c x + d ) d x x − a y d y + b d y = − c d x + d x d x ∫ − a y d y + ∫ b d y = − ∫ c d x + d ∫ d x x ( Integrate ) Further solve the above integral as follows: − a ln y + b y = − c x + d ln x + ln c ln ( x d y a c ) = b y + c x x d y a c = e b y e c x ( x d e − c x ) ( y a e − b y ) = c Consider the F ( x ) and G ( y ) as follows: F ( x ) = x d e − c x G ( y ) = y a e − b y This implies that F ( x ) G ( y ) = c

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 10.4, Problem 7E

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To show: In Lotka-Volterra model if xm<x<xM and x=x1 the equation F(x)G(y)=c0 has exactly two solutions y1 and y2 such that y1<ab<y2.

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Chapter 10 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 10.1 - In Problems 16 write the given nonlinear...Ch. 10.1 - Prob. 2E Ch. 10.1 - In Problems 16 write the given nonlinear...Ch. 10.1 - Prob. 4E Ch. 10.1 - In Problems 16 write the given nonlinear...Ch. 10.1 - In Problems 16 write the given nonlinear...Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - In Problems 716 find all critical points of the...Ch. 10.1 - In Problems 2326 solve the given nonlinear plane...Ch. 10.1 - In Problems 2326 solve the given nonlinear plane...Ch. 10.1 - In Problems 2326 solve the given nonlinear plane...Ch. 10.1 - In Problems 2326 solve the given nonlinear plane...Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.2 - In Problems 916 classify the critical point (0, 0)...Ch. 10.2 - Prob. 10E Ch. 10.2 - Prob. 11E Ch. 10.2 - Prob. 12E Ch. 10.2 - In Problems 916 classify the critical point (0, 0)...Ch. 10.2 - In Problems 916 classify the critical point (0, 0)...Ch. 10.2 - In Problems 916 classify the critical point (0, 0)...Ch. 10.2 - Prob. 16E Ch. 10.2 - Prob. 17E Ch. 10.2 - Determine a condition on the real constant so...Ch. 10.2 - Prob. 19E Ch. 10.2 - Prob. 20E Ch. 10.2 - Prob. 21E Ch. 10.2 - Prob. 22E Ch. 10.2 - Prob. 23E Ch. 10.2 - In Problems 23-26 a nonhomogeneous linear system...Ch. 10.2 - Prob. 25E Ch. 10.2 - Prob. 26E Ch. 10.3 - Prob. 1E Ch. 10.3 - Prob. 2E Ch. 10.3 - Prob. 3E Ch. 10.3 - Prob. 4E Ch. 10.3 - Prob. 5E Ch. 10.3 - In Problems 310, without solving explicitly,...Ch. 10.3 - In Problems 310, without solving explicitly,...Ch. 10.3 - Prob. 8E Ch. 10.3 - Prob. 9E Ch. 10.3 - In Problems 310, without solving explicitly,...Ch. 10.3 - Prob. 11E Ch. 10.3 - In Problems 1120 classify (if possible) each...Ch. 10.3 - Prob. 13E Ch. 10.3 - Prob. 14E Ch. 10.3 - Prob. 15E Ch. 10.3 - Prob. 16E Ch. 10.3 - Prob. 17E Ch. 10.3 - Prob. 18E Ch. 10.3 - Prob. 19E Ch. 10.3 - Prob. 20E Ch. 10.3 - Prob. 21E Ch. 10.3 - Prob. 22E Ch. 10.3 - Prob. 23E Ch. 10.3 - Prob. 24E Ch. 10.3 - Prob. 25E Ch. 10.3 - Prob. 26E Ch. 10.3 - Prob. 27E Ch. 10.3 - Show that the dynamical system x = x + xy y = 1 y...Ch. 10.3 - Prob. 29E Ch. 10.3 - Prob. 30E Ch. 10.3 - Prob. 31E Ch. 10.3 - Prob. 32E Ch. 10.3 - Prob. 33E Ch. 10.3 - Prob. 34E Ch. 10.3 - Prob. 35E Ch. 10.3 - Prob. 36E Ch. 10.3 - When a nonlinear capacitor is present in an...Ch. 10.3 - Prob. 38E Ch. 10.3 - Prob. 39E Ch. 10.4 - Prob. 1E Ch. 10.4 - Prob. 2E Ch. 10.4 - Prob. 3E Ch. 10.4 - Prob. 4E Ch. 10.4 - Prob. 5E Ch. 10.4 - Prob. 6E Ch. 10.4 - Prob. 7E Ch. 10.4 - Prob. 8E Ch. 10.4 - Prob. 9E Ch. 10.4 - Competition Models A competitive interaction is...Ch. 10.4 - Prob. 12E Ch. 10.4 - Prob. 13E Ch. 10.4 - Prob. 14E Ch. 10.4 - Prob. 15E Ch. 10.4 - Additional Mathematical Models Damped Pendulum If...Ch. 10.4 - Prob. 17E Ch. 10.4 - Prob. 18E Ch. 10.4 - Prob. 19E Ch. 10.4 - Prob. 20E Ch. 10.4 - Prob. 21E Ch. 10.4 - Prob. 22E Ch. 10 - Prob. 1RE Ch. 10 - Prob. 2RE Ch. 10 - Prob. 3RE Ch. 10 - Prob. 4RE Ch. 10 - Prob. 5RE Ch. 10 - Prob. 6RE Ch. 10 - Prob. 7RE Ch. 10 - Prob. 8RE Ch. 10 - Prob. 9RE Ch. 10 - Prob. 10RE Ch. 10 - Prob. 11RE Ch. 10 - Discuss the geometric nature of the solutions to...Ch. 10 - Classify the critical point (0, 0) of the given...Ch. 10 - Prob. 14RE Ch. 10 - Prob. 15RE Ch. 10 - Prob. 16RE Ch. 10 - Prob. 17RE Ch. 10 - Prob. 18RE Ch. 10 - Prob. 19RE Ch. 10 - Prob. 20RE

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Finding Local Maxima and Minima by Differentiation; Author: Professor Dave Explains;https://www.youtube.com/watch?v=pvLj1s7SOtk;License: Standard YouTube License, CC-BY

In Lotka-Volterra model if x m < x < x M and x = x 1 the equation F ( x ) G ( y ) = c 0 has exactly two solutions y 1 and y 2 such that y 1 < a b < y 2 .

Solution Summary: The author explains that the plane autonomous system for Lotka-Volterra model is as follows.

Concept explainers

Videos

To show: In Lotka-Volterra model if xm<x<xM and x=x1 the equation F(x)G(y)=c0 has exactly two solutions y1 and y2 such that y1<ab<y2.

Chapter 10 Solutions

In Lotka-Volterra model if x m &lt; x &lt; x M and x = x 1 the equation F ( x ) G ( y ) = c 0 has exactly two solutions y 1 and y 2 such that y 1 &lt; a b &lt; y 2 .

Solution Summary: The author explains that the plane autonomous system for Lotka-Volterra model is as follows.

Concept explainers

Videos

To show: In Lotka-Volterra model if xm<x<xM and x=x1 the equation F(x)G(y)=c0 has exactly two solutions y1 and y2 such that y1<ab<y2.

Chapter 10 Solutions

In Lotka-Volterra model if x m < x < x M and x = x 1 the equation F ( x ) G ( y ) = c 0 has exactly two solutions y 1 and y 2 such that y 1 < a b < y 2 .