Program Explanation Given information: The system is x ′ = [ 0 0 1 − 5 − 1 − 5 4 1 − 2 ] x . Explanation: Consider matrix form of the system of equation as A . A = [ 0 0 1 − 5 − 1 − 5 4 1 − 2 ] The characteristics equation of the coefficient matrix is, [ 0 − λ 0 1 − 5 − 1 − λ − 5 4 1 − 2 − λ ] = 0 ( -λ ) [ − 1 − λ − 5 1 − 2 − λ ] − 0 + ( 1 ) [ − 5 − 1 − λ 4 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, [ 0 − λ 0 1 − 5 − 1 − λ − 5 4 1 − 2 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix Ais v = [ a b c ] T . Substitute − 1 for λ in coefficient matrix, [ 0 + 1 0 1 − 5 − 1 + 1 − 5 4 1 − 2 + 1 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ 1 0 1 − 5 0 − 5 4 1 − 1 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, a 1 + c 1 = 0 − 5 a 1 − 5 c 1 = 0 4 a 1 + b 1 − c 1 = 0 Therefore, the associated Eigen vector is shown below. v 1 = [ a − 5 a − a ] The two Eigen vector is missing, therefore the defect is 2. The other two associated vectors can be obtained by the given formula, v 2 = ( A + I ) v 3 v 1 = ( A + I ) v 2 The value of ( A + I ) 2 can be obtained as, ( A + I ) 2 = [ 1 0 1 − 5 0 − 5 4 1 − 1 ] [ 1 0 1 − 5 0 − 5 4 1 − 1 ] = [ 5 1 0 − 25 − 5 0 − 5 − 1 0 ] The value of ( A + I ) 3 can be obtained as, ( A + I ) 3 = [ 5 1 0 − 25 − 5 0 − 5 − 1 0 ] [ 1 0 1 − 5 0 − 5 4 1 − 1 ] = [ 0 0 0 0 0 0 0 0 0 ] Therefore, any vector v ≠ 0 is a solution of the equation ( A + I ) 3 v = 0

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 14P

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Program Description:Purpose of problem is to find general solution of the system x′=[001−5−1−541−2]x .

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Chapter 5.5, Problem 13P

Chapter 5.5, Problem 15P

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In each of Problems 1 through 8, express the general solution of the given system of equations in terms of real-valued functions:

Problem 5. Let A and k be positive constants. Which of the given functions is a solution to = k(y + A)? O A. y = A-1 + CeAkt O B. y = -A + Cekt O C. y = -A+ Ce-kt O D. y = A+ Cekt O E. y = A+ Ce-kt OF. y = A-+ Ce-Akt

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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 ′ = [ 0 0 1 − 5 − 1 − 5 4 1 − 2 ] 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′=[001−5−1−541−2]x .

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Chapter 5 Solutions