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

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 24P

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

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Chapter 5.2, Problem 23P

Chapter 5.2, Problem 25P

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Exercise 1 Function and Structure [30 pts] Please debug the following program and answer the following questions. There is a cycle in a linked list if some node in the list can be reached again by continuously following the next pointer. #include typedef struct node { int value; struct node *next; } node; int 11_has_cycle (node *first) if (first == node *head = { NULL) return 0; first; while (head->next != NULL) { } if (head first) { return 1; } head = head->next; return 0; void test ll_has_cycle () { int i; node nodes [6]; for (i = 0; i < 6; i++) { nodes [i] .next = NULL; nodes [i].value = i; } nodes [0] .next = &nodes [1]; nodes [1] .next = &nodes [2]; nodes [2] .next = &nodes [3]; nodes [3] .next nodes [4] .next &nodes [4]; NULL; nodes [5] .next = &nodes [0]; printf("1. Checking first list for cycles. \n Function 11_has_cycle says it has s cycle\n\n", 11_has_cycle (&nodes [0])?"a":"no"); printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle says it has %s…

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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 = 2 x 1 + x 2 − x 3 , x ′ 2 = − 4 x 1 − 3 x 2 − x 3 and x ′ 3 = 4 x 1 + 4 x 2 + 2 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=2x1+x2−x3,x′2=−4x1−3x2−x3 and x′3=4x1+4x2+2x3 .

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Chapter 5 Solutions