Program Explanation Given information: The value of A is [ 1 2 2 − 2 ] , f ( t ) = [ 180 t 90 ] , x ( 0 ) = [ 0 0 ] and e A t = 1 5 [ e − 3 t + 4 e 2 t − 2 e − 3 t + 2 e 2 t − 2 e − 3 t + 2 e 2 t 4 e − 3 t + e 2 t ] . Explanation: The initial value problem can be solving by the variation of parameters method as shown below. x ( t ) = e A t x 0 + e A t ∫ 0 t e − A s f ( s ) d s The determinant of e A t can be obtain as, | e A t | = 1 5 [ e − 3 t + 4 e 2 t − 2 e − 3 t + 2 e 2 t − 2 e − 3 t + 2 e 2 t 4 e − 3 t + e 2 t ] = 1 5 | ( e − 3 t + 4 e 2 t ) ( 4 e − 3 t + e 2 t ) − ( − 2 e − 3 t + 2 e 2 t ) ( − 2 e − 3 t + 2 e 2 t ) | = 1 5 ( e − t + 16 e − t + 4 e − t + 4 e − t ) = e − t The determinant of e A t is not 0. Therefore the inverse of the matrix is shown below. e − A t = 1 | e − A t | adj ( e − A t ) = 1 5 e − t [ 4 e − 3 t + e 2 t 2 e − 3 t − 2 e 2 t 2 e − 3 t − 2 e 2 t e − 3 t + 4 e 2 t ] = 1 5 [ 4 e − 2 t + e 3 t 2 e − 2 t − 2 e 3 t 2 e − 2 t − 2 e 3 t e − 2 t + 4 e 3 t ] The value of ∫ 0 t e − A s f ( s ) d s can be obtained as, ∫ 0 t e − A s f ( s ) d s = ∫ 0 t 1 5 [ 4 e − 2 s + e 3 s 2 e − 2 s − 2 e 3 s 2 e

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 5.7, Problem 19P

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Program Description:Purpose of problem is to solve the initial value problem x′=Ax+f(t) and x(a)=xa by using method of variation of parameters.

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Chapter 5.7, Problem 18P

Chapter 5.7, Problem 20P

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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 solve the initial value problem x ′ = A x + f ( t ) and x ( a ) = x a by using method of variation of parameters.

Solution Summary: The author explains that the initial value problem can be solved by the variation of parameters method as shown below.

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Program Description:Purpose of problem is to solve the initial value problem x′=Ax+f(t) and x(a)=xa by using method of variation of parameters.

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Chapter 5 Solutions