Program Explanation Given information: The differential equation is x ′ 1 = 3 x 1 + 2 x 2 + 2 x 3 , x ′ 2 = − 5 x 1 − 3 x 2 − x 3 and x ′ 3 = 5 x 1 + 5 x 2 + 3 x 3 . Explanation: Consider matrix form of the system of equation as A. A x = [ 3 1 1 − 5 − 3 − 1 5 5 3 ] x The characteristics equation of the coefficient matrix is, [ 3 − λ 1 1 − 5 − 3 − λ − 1 5 5 3 − λ ] = 0 ( 3 − λ ) [ − 3 − λ − 1 5 3 − λ ] − 1 [ − 5 − 1 5 3 − λ ] + 1 [ − 5 − 3 − λ 5 5 ] = 0 − λ 3 + 3 λ 2 +4λ − 12 = 0 − λ 2 ( λ − 2 ) + λ ( λ − 2 ) + 6 ( λ − 2 ) = 0 Therefore, the Eigen values of the matrix can be obtained as, ( λ − 2 ) ( − λ 2 + λ+6 ) = 0 ( λ − 2 ) ( − λ 2 + 3 λ − 2 λ + 6 ) = 0 ( λ − 2 ) ( λ − 3 ) ( λ + 2 ) = 0 λ = 3 or λ = − 2 or λ = 2 The Eigen vector of the coefficient matrix A is in the form as, [ 3 − λ 1 1 − 5 − 3 − λ − 1 5 5 3 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix A is v = [ a b c ] T . Substitute 2 for λ in coefficient matrix, [ 3 − 2 1 1 − 5 − 3 − 2 − 1 5 5 3 − 2 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ 1 1 1 − 5 − 5 − 1 5 5 1 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, a 1 + b 1 + c 1 = 0 − 5 a 1 − 5 b 1 − c 1 = 0 5 a 1 + 5 b 1 + c 1 = 0 After solving the equation the value of a 1 , b 1 and c 1 can be obtained as, a 1 = 1 b 1 = − 1 c 1 = 0 Therefore, the associated Eigen vector is shown below. v 1 = [ 1 − 1 0 ] Substitute − 2 for λ in coefficient matrix, [ 3 + 2 1 1 − 5 − 3 + 2 − 1 5 5 3 + 2 ] [ a 2 b 2 c 2 ] = [ 0 0 0 ] [ 5 1 1 − 5 − 1 − 1 5 5 5 ] [ a 2 b 2 c 2 ] = [ 0 0 0 ] The scalar equation are shown below, 5 a 2 + b 2 + c 2 = 0 − 5 a 2 − b 2 − c 2 = 0 5 a 2 + 5 b 2 + 5 c 2 = 0 After solving the equation the value of a 2 , b 2 and c 2 can be obtained as, a 2 = 0 b 2 = 1 c 2 = − 1 Therefore, the associated Eigen vector is shown below

Pearson eText for Differential Equations: Computing and Modeling Tech Update -- Instant Access (Pearson+)

5th Edition

ISBN: 9780137442782

Author: C. Edwards, David Penney

Publisher: PEARSON+

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Chapter 5.2, Problem 23P

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Program Description:Purpose of problem is to obtain the general solution of x′1=3x1+x2+x3,x′2=−5x1−3x2−x3 and x′3=5x1+5x2+3x3 .

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Chapter 5.2, Problem 22P

Chapter 5.2, Problem 24P

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Chapter 5 Solutions

Pearson eText for Differential Equations: Computing and Modeling Tech Update -- Instant Access (Pearson+)

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 x ′ 1 = 3 x 1 + x 2 + x 3 , x ′ 2 = − 5 x 1 − 3 x 2 − x 3 and x ′ 3 = 5 x 1 + 5 x 2 + 3 x 3 .

Solution Summary: The author explains that the purpose of the problem is to obtain the general solution of xprime.

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Program Description:Purpose of problem is to obtain the general solution of x′1=3x1+x2+x3,x′2=−5x1−3x2−x3 and x′3=5x1+5x2+3x3 .

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Chapter 5 Solutions