Program Explanation Given information: The volume of tank 1 is 25 gal and volume of tank 2 is 40 gal . The deferential equation is d x 1 d t = − k 1 x 1 + k 2 x 2 and d x 2 d t = k 1 x 1 − k 2 x 2 The value of r is 10 gal / min , x 1 ( 0 ) = 15 ( l b ) and x 2 ( 0 ) = 0 . Explanation: Consider the rate equation for the tank 1 as follows. d x 1 d t = − r V 1 x 1 + r 2 V 2 x 2 ......... (1) Consider the rate equation for the tank 2 as follows. d x 2 d t = r V 1 x 1 − r V 2 x 2 ......... (2) Substitute 10 for r, 40 gal for V 2 and 25 gal for V 1 in equation (1). d x 1 d t = − 10 25 x 1 + 10 40 x 2 = − 0.4 x 1 + 0.25 x 2 Substitute 10 for r , 40 gal for V 2 and 25 gal for V 1 in equation (2). d x 2 d t = 10 25 x 1 − 10 40 x 2 = 0.4 x 1 − 0.25 x 2 Consider matrix form of the system of equation as A. A = [ − 0.4 0.25 0.4 − 0.25 ] The characteristics equation of the coefficient matrix is, [ − 0.4 − λ 0.25 0.4 − 0.25 − λ ] = 0 ( − 0.4 − λ ) ( − 0.25 − λ ) − 0.1 = 0 λ 2 + 0.65 λ=0 λ = 0 or λ = − 0.65 The Eigen vector of the coefficient matrix A is in the form as, [ − 0.4 − λ 0.25 0.4 − 0.25 − λ ] [ a b ] = [ 0 0 ] The associated Eigen vector with the matrix Ais v = [ a b ] T . Substitute 0 for λ in coefficient matrix, [ − 0.4 − 0 0.25 0.4 − 0.25 − 0 ] [ a 1 b 1 ] = [ 0 0 ] [ 0.4 0.25 0.4 − 0.25 ] [ a 1 b 1 ] = [ 0 0 ] The equation can be obtained as, 0.4 a 1 + 0.25 b 1 = 0 0.4 a 1 − 0.25 b 1 = 0 After solving the equation the value of a 1 and b 1 can be obtained as, a 1 = 5 b 1 = 8 Therefore, the associated Eigen vector is shown below. v 1 = [ 5 8 ] Substitute − 0.6 for λ in coefficient matrix, [ − 0.4 + 0.65 0.25 0.4 − 0.25 + 0.65 ] [ a 2 b 2 ] = [ 0 0 ] [ 0

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ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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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 5.2, Problem 30P

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Program Description:Purpose of problem is to obtain the solution of differential equation dx1dt=−k1x1+k2x2 and dx2dt=k1x1−k2x2 and also draw a graph of x1(t) and x2(t) .

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Chapter 5.2, Problem 29P

Chapter 5.2, Problem 31P

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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 solution of differential equation d x 1 d t = − k 1 x 1 + k 2 x 2 and d x 2 d t = k 1 x 1 − k 2 x 2 and also draw a graph of x 1 ( t ) and x 2 ( t ) .

Solution Summary: The author explains that the purpose of the problem is to obtain the solution of differential equation dx_1.

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Program Description:Purpose of problem is to obtain the solution of differential equation dx1dt=−k1x1+k2x2 and dx2dt=k1x1−k2x2 and also draw a graph of x1(t) and x2(t) .

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Chapter 5 Solutions