Program Explanation Given information: The system is x ′ = [ 1 − 4 0 − 2 0 1 0 0 6 − 12 − 1 − 6 0 − 4 0 − 1 ] x . Explanation: Consider matrix form of the system of equation as A . A = [ 1 − 4 0 − 2 0 1 0 0 6 − 12 − 1 − 6 0 − 4 0 − 1 ] The characteristics equation of the coefficient matrix is, [ 1 − λ − 4 0 − 2 0 1 − λ 0 0 6 − 12 − 1 − λ − 6 0 − 4 0 − 1 − λ ] = 0 ( λ 2 − 1 ) 4 = 0 λ = 1 , 1 , − 1 and -1 The obtain Eigen value has multiplicity 4. The Eigen vector of the coefficient matrix A is in the form as, [ 1 − λ − 4 0 − 2 0 1 − λ 0 0 6 − 12 − 1 − λ − 6 0 − 4 0 − 1 − λ ] [ a b c d ] = [ 0 0 0 0 ] The associated Eigen vector with the matrix A is v = [ a b c d ] T . Substitute 1 for λ in coefficient matrix, [ 1 − 1 − 4 0 − 2 0 1 − 1 0 0 6 − 12 − 1 − 1 − 6 0 − 4 0 − 1 − 1 ] [ a b c d ] = [ 0 0 0 0 ] [ 0 − 4 0 − 2 0 0 0 0 6 − 12 − 2 − 6 0 − 4 0 − 2 ] [ a b c d ] = [ 0 0 0 0 ] The scalar equation are shown below, − 4 b − 2 d = 0 6 a − 12 b − 2 c − 6 d = 0 − 4 b − 2 d = 0 Therefore, the associated Eigen vector is obtained as: u 1 = [ 1 0 3 0 ] The value of ( A − I ) 2 can be obtained as, ( A − I ) 2 = [ 0 − 4 0 − 2 0 0 0 0 6 − 12 − 2 − 6 0 − 4 0 − 2 ] [ 0 − 4 0 − 2 0 0 0 0 6 − 12 − 2 − 6 0 − 4 0 − 2 ] = [ 0 8 0 4 0 0 0 0 − 12 24 4 12 0 8 0 4 ] Therefore, any vector u ≠ 0 is a solution of the equation ( A − I ) 2 v = 0

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Problems on Growing Functions

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Question

Chapter 5.5, Problem 19P

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[1−40−201006−12−1−60−40−1]x .

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Chapter 5.5, Problem 18P

Chapter 5.5, Problem 20P

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Chapter 5 Solutions

EBK DIFFERENTIAL EQUATIONS

Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

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Program Description: Purpose of problem is to obtain the general solution of the system x ′ = [ 1 − 4 0 − 2 0 1 0 0 6 − 12 − 1 − 6 0 − 4 0 − 1 ] x .

Solution Summary: The author explains that the problem is to obtain the general solution of the system of equation as A.

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[1−40−201006−12−1−60−40−1]x .

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Chapter 5 Solutions