Program Explanation Given information: The system is x ′ = [ − 1 0 1 0 1 − 4 1 1 − 3 ] x . Explanation: Consider matrix form of the system of equation as A . A = [ − 1 0 1 0 1 − 4 1 1 − 3 ] The characteristics equation of the coefficient matrix is, [ − 1 − λ 0 1 0 1 − λ − 4 1 1 − 3 − λ ] = 0 ( − 1 − λ ) [ 1 − λ − 4 1 − 3 − λ ] − 0 + ( 1 ) [ 0 1 − λ 1 1 ] = 0 λ 3 + 3 λ 2 + 3 λ+1 = 0 Further solve the above equation as shown below. λ 3 + 3 λ 2 + 3 λ+1 = 0 ( λ + 1 ) 3 = 0 λ = − 1 , − 1 , and − 1 The Eigen vector of the coefficient matrix A is in the form as, [ − 1 − λ 0 1 0 − 1 − λ 1 1 − 1 1 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix A is v = [ a b c ] T . Substitute − 1 for λ in coefficient matrix, [ − 1 + 1 0 1 0 1 + 1 − 4 1 1 − 3 + 1 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ 0 0 1 0 2 − 4 1 1 − 2 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, c 1 = 0 2 b 1 − 4 c 1 = 0 b 1 − 2 c 1 = 0 If a 1 is the arbitrary number such as 1 with condition v ≠ 0 then the associated Eigen vector is shown below: v 1 = [ 1 0 0 ] Similarly if a 1 is 0 then the b must be non zero then the associated Eigen vector is shown below: v 2 = [ 0 2 1 ] The defect is 2

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

5th Edition

ISBN: 9780321816252

Author: C. Henry Edwards, David E. Penney, David Calvis

Publisher: PEARSON

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Chapter 5.5, Problem 13P

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Program Description:Purpose of problem is to find general solution of the system x′=[−10101−411−3]x .

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Chapter 5.5, Problem 12P

Chapter 5.5, Problem 14P

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Chapter 5 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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 find general solution of the system x ′ = [ − 1 0 1 0 1 − 4 1 1 − 3 ] x .

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

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Program Description:Purpose of problem is to find general solution of the system x′=[−10101−411−3]x .

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Chapter 5 Solutions