Program Explanation Given information: The physical parameters isgiven as, m 1 = m 2 = c 1 = c 2 = c = k = 1 The initial conditions are, x 1 ( 0 ) = x 2 ( 0 ) = 0 and x ′ 1 ( 0 ) = x ′ 2 ( 0 ) = v 0 Explanation: The equation of motion of cars is defined as, m 1 x ″ 1 = k ( x 2 − x 1 ) − c 1 x ′ 1 − c ( x ′ 1 − x ′ 2 ) m 2 x ″ 2 = k ( x 1 − x 2 ) − c 2 x ′ 2 − c ( x ′ 2 − x ′ 1 ) The characteristic equation for the above equation of motion is shown below. A = [ 0 0 1 0 0 0 0 1 − k k − ( c 1 + c 2 ) c k − k c − ( c 1 + c 2 ) ] Substitute the value of physical parameters to obtain the characteristic equation as, A = [ 0 0 1 0 0 0 0 1 − 1 1 − 2 1 1 − 1 1 − 2 ] The characteristics equation of the coefficient matrix is, [ 0 − λ 0 1 0 0 0 − λ 0 1 − 1 1 − 2 − λ 1 1 − 1 1 − 2 − λ ] = 0 λ ( λ + 1 ) 2 ( λ + 2 ) = 0 λ = 0 , − 1 , − 1 and -2 The obtain Eigen value − 1 has multiplicity 2. The Eigen vector of the coefficient matrix A is in the form as, [ 0 − λ 0 1 0 0 0 − λ 0 1 − 1 1 − 2 − λ 1 1 − 1 1 − 2 − λ ] [ a b c d ] = [ 0 0 0 0 ] The associated Eigen vector with the matrix A is v = [ a b c d ] T . Substitute 0 for λ in coefficient matrix, [ 0 − 0 0 1 0 0 0 − 0 0 1 − 1 1 − 2 − 0 1 1 − 1 1 − 2 − 0 ] [ a 1 b 1 c 1 d 1 ] = [ 0 0 0 0 ] [ 0 0 1 0 0 0 0 1 − 1 1 − 2 1 1 − 1 1 − 2 ] [ a 1 b 1 c 1 d 1 ] = [ 0 0 0 0 ] The scalar equation are shown below, c 1 = 0 d 1 = 0 − a 1 + b 1 − 2 c 1 + d 1 = 0 a 1 − b 1 + c 1 − 2 d 1 = 0 Therefore, the associated Eigen vector is obtained as: v 0 = [ 1 1 0 0 ] Substitute − 1 for λ in coefficient matrix, [ 0 + 1 0 1 0 0 0 + 1 0 1 − 1 1 − 2 + 1 1 1 − 1 1 − 2 + 1 ] [ a 2 b 2 c 2 d 2 ] = [ 0 0 0 0 ] [ 1 0 1 0 0 1 0 1 − 1 1 − 1 1 1 − 1 1 − 1 ] [ a 2 b 2 c 2 d 2 ] = [ 0 0 0 0 ] The scalar equation are shown below, a 2 + c 2 = 0 b 2 + d 2 = 0 − a 2 + b 2 − c 2 + d 2 = 0 a 2 − b 2 + c 2 − d 2 = 0 Therefore, the associated Eigen vector is obtained as: u 1 = [ − 1 0 1 0 ] The other associated Eigen vectors is obtained as, u 2 = [ 0 − 1 0 1 ] Substitute 2 for λ in coefficient matrix, [ 0 + 2 0 1 0 0 0 + 2 0 1 − 1 1 − 2 + 2 1 1 − 1 1 − 2 + 2 ] [ a 3 b 3 c 3 d 3 ] = [ 0 0 0 0 ] [ 2 0 1 0 0 2 0 1 − 1 1 0 1 1 − 1 1 0 ] [ a 3 b 3 c 3 d 3 ] = [ 0 0 0 0 ] The scalar equation are shown below, 2 a 3 + c 3 = 0 2 b 3 + d 3 = 0 − a 3 + b 3 + d 3 = 0 a 3 − b 3 + c 3 = 0 The

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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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 5.5, Problem 35P

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Program Description:Purpose of problem is to obtain the position function x1(t) and x2(t) of the railway cars and also determine the distance travel by cars before stop.

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Chapter 5.5, Problem 34P

Chapter 5.5, Problem 36P

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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 position function x 1 ( t ) and x 2 ( t ) of the railway cars and also determine the distance travel by cars before stop.

Solution Summary: The author explains that the equation of motion of cars is defined as, lm_1x

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Program Description:Purpose of problem is to obtain the position function x1(t) and x2(t) of the railway cars and also determine the distance travel by cars before stop.

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Chapter 5 Solutions