Program Explanation Given information: The differential equation shown below: x ′ 1 = 4 x 1 + x 2 + x 3 + 7 x 4 x ′ 2 = x 1 + 4 x 2 + 10 x 3 + x 4 x ′ 3 = x 1 + 10 x 2 + 4 x 3 + x 4 x ′ 4 = 7 x 1 + x 2 + x 3 + 4 x 4 The Eigen values of the coefficient matrix is − 3 , − 6 , − 10 , and 15 . The initial condition is shown below. x 1 ( 0 ) = 3 , x 2 ( 0 ) = x 3 ( 0 ) = 1 and x 4 ( 0 ) = 3 Explanation: Consider matrix form of the system of equation as A. A = [ 4 1 1 7 1 4 10 1 1 10 4 1 7 1 1 4 ] The Eigen vector of the coefficient matrix A is in the form as, [ 4 − λ 1 1 7 1 4 − λ 10 1 1 10 4 − λ 1 7 1 1 4 − λ ] [ a b c d ] = [ 0 0 0 0 ] The associated Eigen vector with the matrix Ais v = [ a b c d ] T . Substitute − 3 for λ in coefficient matrix, [ 4 + 3 1 1 7 1 4 + 3 10 1 1 10 4 + 3 1 7 1 1 4 + 3 ] [ a 1 b 1 c 1 d 1 ] = [ 0 0 0 0 ] [ 7 1 1 7 1 7 10 1 1 10 7 1 7 1 1 7 ] [ a 1 b 1 c 1 d 1 ] = [ 0 0 0 0 ] The scalar equation are shown below, 7 a 1 + b 1 + c 1 + 7 d 1 = 0 a 1 + 7 b 1 + 10 c 1 + d 1 = 0 a 1 + 10 b 1 + 7 c 1 + 7 d 1 = 0 7 a 1 + b 1 + c 1 + 7 d 1 = 0 After solving the equation the value of a 1 , b 1 , c 1 and d 1 can be obtained as, a 1 = 1 b 1 = 0 c 1 = 0 d 1 = − 1 Therefore, the associated Eigen vector is shown below. v 1 = [ 1 0 0 − 1 ] Substitute − 6 for λ in coefficient matrix, [ 4 + 6 1 1 7 1 4 + 6 10 1 1 10 4 + 6 1 7 1 1 4 + 6 ] [ a 2 b 2 c 2 d 2 ] = [ 0 0 0 0 ] [ 10 1 1 7 1 10 10 1 1 10 10 1 7 1 1 10 ] [ a 2 b 2 c 2 d 2 ] = [ 0 0 0 0 ] The scalar equation are shown below, 10 a 2 + b 2 + c 2 + 7 d 2 = 0 a 2 + 10 b 2 + 10 c 2 + d 2 = 0 a 2 + 10 b 2 + 10 c 2 + d 2 = 0 7 a 2 + b 2 + c 2 + 10 d 2 = 0 After solving the equation the value of a 2 , b 2 , c 2 and d 2 can be obtained as, a 2 = 0 b 2 = 1 c 2 = − 1 d 2 = 0 Therefore, the associated Eigen vector is shown below. v 2 = [ 0 1 − 1 0 ] Substitute 10 for λ in coefficient matrix, [ 4 − 10 1 1 7 1 4 − 10 10 1 1 10 4 − 10 1 7 1 1 4 − 10 ] [ a 3 b 3 c 3 d 3 ] = [ 0 0 0 0 ] [ − 6 1 1 7 1 − 6 10 1 1 10 − 6 1 7 1 1 − 6 ] [ a 3 b 3 c 3 d 3 ] = [ 0 0 0 0 ] The scalar equation are shown below, − 6 a 3 + b 3 + c 3 + 7 d 3 = 0 a 3 − 6 b 3 + 10 c 3 + d 4 = 0 a 3 + 10 b 3 − 6 c 3 + d 3 = 0 7 a 3 + b 3 + c 3 − 6 d 3 = 0 After solving the equation the value of a 3 , b 3 , c 3 and d 3 can be obtained as, a 3 = 2 b 3 = − 1 c 3 = − 1 d 3 = 2 Therefore, the associated Eigen vector is shown below

5th Edition

ISBN: 8220100802454

Author: Calvis

Publisher: PEARSON

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Adjacency matrix representation

An adjacency matrix is a square matrix that represents a finite graph. The matrix denotes whether a particular pair of nodes in a graph is adjacent or not. In brief, an adjacency matrix is an array of size N x N, where N is the number of vertices in the …

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Chapter 5.2, Problem 41P

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Program Description:Purpose of problem is to obtain the particular solution of the coefficient matrix A.

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Chapter 5.2, Problem 40P

Chapter 5.2, Problem 42P

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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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(53) by substituting the expressions...Ch. 5.3 - Prob. 35P Ch. 5.3 - Prob. 36P Ch. 5.3 - Prob. 37P Ch. 5.3 - Prob. 38P Ch. 5.3 - Prob. 39P Ch. 5.3 - Prob. 40P Ch. 5.4 - Prob. 1P Ch. 5.4 - Prob. 2P Ch. 5.4 - Prob. 3P Ch. 5.4 - Prob. 4P Ch. 5.4 - Prob. 5P Ch. 5.4 - Prob. 6P Ch. 5.4 - Prob. 7P Ch. 5.4 - Prob. 8P Ch. 5.4 - Prob. 9P Ch. 5.4 - Prob. 10P Ch. 5.4 - Prob. 11P Ch. 5.4 - Prob. 12P Ch. 5.4 - Prob. 13P Ch. 5.4 - Prob. 14P Ch. 5.4 - Prob. 15P Ch. 5.4 - Prob. 16P Ch. 5.4 - Prob. 17P Ch. 5.4 - Prob. 18P Ch. 5.4 - Prob. 19P Ch. 5.4 - Prob. 20P Ch. 5.4 - Prob. 21P Ch. 5.4 - Prob. 22P Ch. 5.4 - Prob. 23P Ch. 5.4 - Prob. 24P Ch. 5.4 - Prob. 25P Ch. 5.4 - Prob. 26P Ch. 5.4 - Prob. 27P Ch. 5.4 - Prob. 28P Ch. 5.4 - Prob. 29P Ch. 5.5 - Prob. 1P Ch. 5.5 - Prob. 2P Ch. 5.5 - Prob. 3P Ch. 5.5 - Prob. 4P Ch. 5.5 - Prob. 5P Ch. 5.5 - Prob. 6P Ch. 5.5 - Prob. 7P Ch. 5.5 - Prob. 8P Ch. 5.5 - Prob. 9P Ch. 5.5 - Prob. 10P Ch. 5.5 - Prob. 11P Ch. 5.5 - Prob. 12P Ch. 5.5 - Prob. 13P Ch. 5.5 - Prob. 14P Ch. 5.5 - Prob. 15P Ch. 5.5 - Prob. 16P Ch. 5.5 - Prob. 17P Ch. 5.5 - Prob. 18P Ch. 5.5 - Prob. 19P Ch. 5.5 - Prob. 20P Ch. 5.5 - Prob. 21P Ch. 5.5 - Prob. 22P Ch. 5.5 - Prob. 23P Ch. 5.5 - Prob. 24P Ch. 5.5 - Prob. 25P Ch. 5.5 - Prob. 26P Ch. 5.5 - Prob. 27P Ch. 5.5 - Prob. 28P Ch. 5.5 - Prob. 29P Ch. 5.5 - Prob. 30P Ch. 5.5 - Prob. 31P Ch. 5.5 - Prob. 32P Ch. 5.5 - Prob. 33P Ch. 5.5 - Prob. 34P Ch. 5.5 - Prob. 35P Ch. 5.5 - Prob. 36P Ch. 5.6 - Prob. 1P Ch. 5.6 - Prob. 2P Ch. 5.6 - Prob. 3P Ch. 5.6 - Prob. 4P Ch. 5.6 - Prob. 5P Ch. 5.6 - Prob. 6P Ch. 5.6 - Prob. 7P Ch. 5.6 - Prob. 8P Ch. 5.6 - Prob. 9P Ch. 5.6 - Prob. 10P Ch. 5.6 - Prob. 11P Ch. 5.6 - Prob. 12P Ch. 5.6 - Prob. 13P Ch. 5.6 - Prob. 14P Ch. 5.6 - Prob. 15P Ch. 5.6 - Prob. 16P Ch. 5.6 - Prob. 17P Ch. 5.6 - Prob. 18P Ch. 5.6 - Prob. 19P Ch. 5.6 - Prob. 20P Ch. 5.6 - Prob. 21P Ch. 5.6 - Prob. 22P Ch. 5.6 - Prob. 23P Ch. 5.6 - Prob. 24P Ch. 5.6 - Prob. 25P Ch. 5.6 - Prob. 26P Ch. 5.6 - Prob. 27P Ch. 5.6 - Prob. 28P Ch. 5.6 - Prob. 29P Ch. 5.6 - Prob. 30P Ch. 5.6 - Prob. 31P Ch. 5.6 - Prob. 32P Ch. 5.6 - Prob. 33P Ch. 5.6 - Prob. 34P Ch. 5.6 - Prob. 35P Ch. 5.6 - Prob. 36P Ch. 5.6 - Prob. 37P Ch. 5.6 - Prob. 38P Ch. 5.6 - Prob. 39P Ch. 5.6 - Prob. 40P Ch. 5.7 - Prob. 1P Ch. 5.7 - Prob. 2P Ch. 5.7 - Prob. 3P Ch. 5.7 - Prob. 4P Ch. 5.7 - Prob. 5P Ch. 5.7 - Prob. 6P Ch. 5.7 - Prob. 7P Ch. 5.7 - Prob. 8P Ch. 5.7 - Prob. 9P Ch. 5.7 - Prob. 10P Ch. 5.7 - Prob. 11P Ch. 5.7 - Prob. 12P Ch. 5.7 - Prob. 13P Ch. 5.7 - Prob. 14P Ch. 5.7 - Prob. 15P Ch. 5.7 - Prob. 16P Ch. 5.7 - Prob. 17P Ch. 5.7 - Prob. 18P Ch. 5.7 - Prob. 19P Ch. 5.7 - Prob. 20P Ch. 5.7 - Prob. 21P Ch. 5.7 - Prob. 22P Ch. 5.7 - Prob. 23P Ch. 5.7 - Prob. 24P Ch. 5.7 - Prob. 25P Ch. 5.7 - Prob. 26P Ch. 5.7 - Prob. 27P Ch. 5.7 - Prob. 28P Ch. 5.7 - Prob. 29P Ch. 5.7 - Prob. 30P Ch. 5.7 - Prob. 31P Ch. 5.7 - Prob. 32P Ch. 5.7 - Prob. 33P Ch. 5.7 - Prob. 34P

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Program Description: Purpose of problem is to obtain the particular solution of the coefficient matrix A.

Solution Summary: The author explains that the problem is to obtain the particular solution of the coefficient matrix A.

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Program Description:Purpose of problem is to obtain the particular solution of the coefficient matrix A.

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Chapter 5 Solutions