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

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 5.2, Problem 21P

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Program Description:Purpose of problem is to obtain the general solution of x′1=5x1−6x3,x′2=2x1−x2−2x3 and x′3=4x1−2x2−4x3 .

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Chapter 5.2, Problem 20P

Chapter 5.2, Problem 22P

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In each of Problems 5 and 6 the coefficient matrix has a zero eigenvalue. As a result, the pattern of trajectories is different from those in the examples in the text. For each system: Ga. Draw a direction field. b. Find the general solution of the given system of equations. G c. Draw a few of the trajectories. 4 -3 8 -6 5. x' = X

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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

Ch. 5.1 - Prob. 11P Ch. 5.1 - Prob. 12P Ch. 5.1 - Prob. 13P Ch. 5.1 - Prob. 14P Ch. 5.1 - Prob. 15P Ch. 5.1 - Prob. 16P Ch. 5.1 - Prob. 17P Ch. 5.1 - Prob. 18P Ch. 5.1 - Prob. 19P Ch. 5.1 - Prob. 20P Ch. 5.1 - Prob. 21P Ch. 5.1 - Prob. 22P Ch. 5.1 - Prob. 23P Ch. 5.1 - Prob. 24P Ch. 5.1 - Prob. 25P Ch. 5.1 - Prob. 26P Ch. 5.1 - Prob. 27P Ch. 5.1 - Prob. 28P Ch. 5.1 - Prob. 29P Ch. 5.1 - Prob. 30P Ch. 5.1 - Prob. 31P Ch. 5.1 - Prob. 32P Ch. 5.1 - Prob. 33P Ch. 5.1 - Prob. 34P Ch. 5.1 - Prob. 35P Ch. 5.1 - Prob. 36P Ch. 5.1 - Prob. 37P Ch. 5.1 - Prob. 38P Ch. 5.1 - Prob. 39P Ch. 5.1 - Prob. 40P Ch. 5.1 - Prob. 41P Ch. 5.1 - Prob. 42P Ch. 5.1 - Prob. 43P Ch. 5.1 - Prob. 44P Ch. 5.1 - Prob. 45P Ch. 5.2 - Prob. 1P Ch. 5.2 - Prob. 2P Ch. 5.2 - Prob. 3P Ch. 5.2 - Prob. 4P Ch. 5.2 - Prob. 5P Ch. 5.2 - Prob. 6P Ch. 5.2 - Prob. 7P Ch. 5.2 - Prob. 8P Ch. 5.2 - Prob. 9P Ch. 5.2 - Prob. 10P Ch. 5.2 - Prob. 11P Ch. 5.2 - Prob. 12P Ch. 5.2 - Prob. 13P Ch. 5.2 - Prob. 14P Ch. 5.2 - Prob. 15P Ch. 5.2 - Prob. 16P Ch. 5.2 - Prob. 17P Ch. 5.2 - Prob. 18P Ch. 5.2 - Prob. 19P Ch. 5.2 - Prob. 20P Ch. 5.2 - Prob. 21P Ch. 5.2 - Prob. 22P Ch. 5.2 - Prob. 23P Ch. 5.2 - Prob. 24P Ch. 5.2 - Prob. 25P Ch. 5.2 - Prob. 26P Ch. 5.2 - Prob. 27P Ch. 5.2 - Prob. 28P Ch. 5.2 - Prob. 29P Ch. 5.2 - Prob. 30P Ch. 5.2 - Prob. 31P Ch. 5.2 - Prob. 32P Ch. 5.2 - Prob. 33P Ch. 5.2 - Prob. 34P Ch. 5.2 - Prob. 35P Ch. 5.2 - Prob. 36P Ch. 5.2 - Prob. 37P Ch. 5.2 - Prob. 38P Ch. 5.2 - Prob. 39P Ch. 5.2 - Prob. 40P Ch. 5.2 - Prob. 41P Ch. 5.2 - Prob. 42P Ch. 5.2 - Prob. 43P Ch. 5.2 - Prob. 44P Ch. 5.2 - Prob. 45P Ch. 5.2 - Prob. 46P Ch. 5.2 - Prob. 47P Ch. 5.2 - Prob. 48P Ch. 5.2 - Prob. 49P Ch. 5.2 - Prob. 50P Ch. 5.3 - Prob. 1P Ch. 5.3 - Prob. 2P Ch. 5.3 - Prob. 3P Ch. 5.3 - Prob. 4P Ch. 5.3 - Prob. 5P Ch. 5.3 - Prob. 6P Ch. 5.3 - Prob. 7P Ch. 5.3 - Prob. 8P Ch. 5.3 - Prob. 9P Ch. 5.3 - Prob. 10P Ch. 5.3 - Prob. 11P Ch. 5.3 - Prob. 12P Ch. 5.3 - Prob. 13P Ch. 5.3 - Prob. 14P Ch. 5.3 - Prob. 15P Ch. 5.3 - Prob. 16P Ch. 5.3 - Prob. 17P Ch. 5.3 - Prob. 18P Ch. 5.3 - Prob. 19P Ch. 5.3 - Prob. 20P Ch. 5.3 - Prob. 21P Ch. 5.3 - Prob. 22P Ch. 5.3 - Prob. 23P Ch. 5.3 - Prob. 24P Ch. 5.3 - Prob. 25P Ch. 5.3 - Prob. 26P Ch. 5.3 - Prob. 27P Ch. 5.3 - Prob. 28P Ch. 5.3 - Prob. 29P Ch. 5.3 - Prob. 30P Ch. 5.3 - Prob. 31P Ch. 5.3 - Prob. 32P Ch. 5.3 - Prob. 33P Ch. 5.3 - Verify Eq. 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Program Description: Purpose of problem is to obtain the general solution of x ′ 1 = 5 x 1 − 6 x 3 , x ′ 2 = 2 x 1 − x 2 − 2 x 3 and x ′ 3 = 4 x 1 − 2 x 2 − 4 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=5x1−6x3,x′2=2x1−x2−2x3 and x′3=4x1−2x2−4x3 .

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Chapter 5 Solutions