Explanation Given: The given differential equation is, d x d t = − 0.1 x + 0.02 x y d y d t = 0.2 y − 0.025 x y Where the population of predators are represented by x ( t ) and the population of prey are represented by y ( t ) . And, the given initial conditions are x ( 0 ) = 6 and y ( 0 ) = 6 . Calculation: Consider the following equation to calculate equilibrium points, d x d t = 0 d y d t = 0 Substitute 0 for d x d t and d y d t in the given equations. − 0.1 x + 0.02 x y = 0 … … ( 1 ) 0.2 y − 0.025 x y = 0 … … ( 2 ) Simplify equation (1) as, − x + 0.2 x y = 0 ( 0.2 y − 1 ) x = 0 y = 5 Simplify equation (2) as, 2 y − 0.25 x y = 0 ( 2 − 0.25 x ) y = 0 x = 8 Therefore, the equilibrium points are ( 0 , 0 ) and ( 8 , 5 ) . Consider the following equation to linearize the given equations. { f ( x , y ) = − 0.1 x + 0.02 x y g ( x , y ) = 0.2 y − 0.025 x y } Therefore, the partial derivatives will be, { f x = − 0.1 + 0.02 y g x = − 0.025 y } and { f y = 0.02 x g y = 0.2 − 0.025 x } Thus, the Jacobean matrix is given by, ( f x f y g x g y ) = ( − 0.1 + 0.02 y 0.02 x − 0.025 y 0.2 − 0.025 x ) … … ( 3 ) Substitute ( 0 , 0 ) for ( x , y ) in equation (3). ( f x f y g x g y ) ( 0 , 0 ) = ( − 0.1 + 0.02 ( 0 ) 0.02 ( 0 ) − 0.025 ( 0 ) 0.2 − 0.025 ( 0 ) ) ( f x f y g x g y ) ( 0 , 0 ) = ( − 0.1 0 0 0

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.9, Problem 11E

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To graph: The solutions x(t) and y(t) and calculate the time at which both the populations of predators and prey are first equal and approximate the period of each population.

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Chapter 2.9, Problem 10E

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To graph: The solutions x ( t ) and y ( t ) and calculate the time at which both the populations of predators and prey are first equal and approximate the period of each population.

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To graph: The solutions x(t) and y(t) and calculate the time at which both the populations of predators and prey are first equal and approximate the period of each population.

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Chapter 2 Solutions