Program Explanation Given information: The differential equation is x ′ 1 = 5 x 1 + x 2 + 3 x 3 , x ′ 2 = x 1 + 7 x 2 + x 3 and x ′ 3 = 3 x 1 + x 2 + 5 x 3 . Explanation: Consider matrix form of the system of equation as A. A x = [ 5 1 3 1 7 1 3 1 5 ] x The characteristics equation of the coefficient matrix is, [ 5 − λ 1 3 1 7 − λ 1 3 1 5 − λ ] = 0 ( 5 − λ ) [ 7 − λ 1 1 5 − λ ] − 1 [ 1 1 3 5 − λ ] + 3 [ 1 7 − λ 3 1 ] = 0 − λ 3 + 17 λ 2 − 84 λ+108 = 0 − λ 2 ( λ − 2 ) + 15 λ ( λ − 2 ) − 54 ( λ − 2 ) = 0 Therefore, the Eigen values of the matrix can be obtained as, ( λ − 2 ) ( − λ 2 + 9 λ+6λ − 54 ) = 0 ( λ − 2 ) ( λ − 6 ) ( λ − 9 ) = 0 λ = 2 or λ = 6 or λ = 9 The Eigen vector of the coefficient matrix A is in the form as, [ 5 − λ 1 3 1 7 − λ 1 3 1 5 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix Ais v = [ a b c ] T . Substitute 9 for λ in coefficient matrix, [ 5 − 9 1 3 1 7 − 9 1 3 1 5 − 9 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ − 4 1 3 1 − 2 1 3 1 − 4 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, − 4 a 1 + b 1 + 3 c 1 = 0 a 1 − 2 b 1 + c 1 = 0 3 a 1 + b 1 − 4 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 = 1 Therefore, the associated Eigen vector is shown below. v 1 = [ 1 1 1 ] Substitute 6 for λ in coefficient matrix, [ 5 − 6 1 3 1 7 − 6 1 3 1 5 − 6 ] [ a 2 b 2 c 2 ] = [ 0 0 0 ] [ − 1 1 3 1 1 1 3 1 − 1 ] [ a 2 b 2 c 2 ] = [ 0 0 0 ] The scalar equation are shown below, − a 2 + b 2 + 3 c 2 = 0 a 2 + b 2 + c 2 = 0 3 a 2 + b 2 + c 2 = 0 After solving the equation the value of a 2 , b 2 and c 2 can be obtained as, a 2 = 1 b 2 = − 2 c 2 = 1 Therefore, the associated Eigen vector is shown below

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Problems on Growing Functions

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Chapter 5.2, Problem 20P

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Program Description:Purpose of problem is to obtain the general solution of x′1=5x1+x2+3x3,x′2=x1+7x2+x3 and x′3=3x1+x2+5x3 .

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Chapter 5.2, Problem 19P

Chapter 5.2, Problem 21P

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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 + x 2 + 3 x 3 , x ′ 2 = x 1 + 7 x 2 + x 3 and x ′ 3 = 3 x 1 + x 2 + 5 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=5x1+x2+3x3,x′2=x1+7x2+x3 and x′3=3x1+x2+5x3 .

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Chapter 5 Solutions