Program Explanation Given information: The differential equation is x ′ 1 = 5 x 1 + 5 x 2 + 2 x 3 , x ′ 2 = − 6 x 1 − 6 x 2 − 5 x 3 and x ′ 3 = 6 x 1 + 6 x 2 + 5 x 3 . Explanation: Consider matrix form of the system of equation as A. A x = [ 5 5 2 − 6 − 6 − 5 6 6 5 ] x The characteristics equation of the coefficient matrix is, [ 5 − λ 5 2 − 6 − 6 − λ − 5 6 6 5 − λ ] = 0 ( 5 − λ ) [ − 6 − λ − 5 6 5 − λ ] − 5 [ − 6 − 5 6 5 − λ ] + 1 [ − 6 − 6 − λ 6 6 ] = 0 λ 3 − λ 2 +5λ 2 +5λ + 18 λ = 0 λ 3 − 4λ 2 + 13 λ = 0 Therefore, the Eigen values of the matrix can be obtained as, λ ( λ 2 − 4 λ+13 ) = 0 ( λ ) ( λ + 2 + 3 i ) ( λ+ 2 − 3 i ) = 0 λ = 0 or λ = 2 + 3 i or λ = 2 − 3 i The Eigen vector of the coefficient matrix A is in the form as, [ 5 − λ 5 2 − 6 − 6 − λ − 5 6 6 5 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix Ais v = [ a b c ] T . Substitute 0 for λ in coefficient matrix, [ 5 − 0 5 2 − 6 − 6 − 0 − 5 6 6 5 − 0 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ 5 5 2 − 6 − 6 − 5 6 6 5 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, 5 a 1 + 5 b 1 + 2 c 1 = 0 − 6 a 1 − 6 b 1 − 5 c 1 = 0 6 a 1 + 6 b 1 + 5 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 + 3 i for λ in coefficient matrix, [ 5 − 2 − 3 i 5 2 − 6 − 6 − 2 − 3 i − 5 6 6 5 − 2 − 3 i ] [ a 2 b 2 c 2 ] = [ 0 0 0 ] [ 3 − 3 i 5 2 − 6 − 8 − 3 i − 5 6 6 3 − 3 i ] [ a 2 b 2 c 2 ] = [ 0 0 0 ] The scalar equation are shown below, ( 3 − 3 i ) a 2 + 5 b 2 + 2 c 2 = 0 − 6 a 2 − ( 8 + 3 i ) b 2 − 5 c 2 = 0 6 a 2 + 6 b 2 + ( 3 − 2 i ) c 2 = 0 After solving the equation the value of a 2 , b 2 and c 2 can be obtained as, a 2 = 1 + i b 2 = − 2 c 2 = 2 Therefore, the associated Eigen vector is shown below

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 5.2, Problem 25P

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Program Description:Purpose of problem is to obtain the general solution of x′1=5x1+5x2+2x3,x′2=−6x1−6x2−5x3 and x′3=6x1+6x2+5x3 by Eigen values method.

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Chapter 5.2, Problem 24P

Chapter 5.2, Problem 26P

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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 x ′ 1 = 5 x 1 + 5 x 2 + 2 x 3 , x ′ 2 = − 6 x 1 − 6 x 2 − 5 x 3 and x ′ 3 = 6 x 1 + 6 x 2 + 5 x 3 by Eigen values method.

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

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Program Description:Purpose of problem is to obtain the general solution of x′1=5x1+5x2+2x3,x′2=−6x1−6x2−5x3 and x′3=6x1+6x2+5x3 by Eigen values method.

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Chapter 5 Solutions