Program Explanation Given information: The volume of tank 1 is 50 gal and volume of tank 2 is 25 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, 25 gal for V 2 and 50 gal for V 1 in equation (1). d x 1 d t = − 10 50 x 1 + 10 25 x 2 = − 0.2 x 1 + 0.4 x 2 Substitute 10 for r , 25 gal for V 2 and 50 gal for V 1 in equation (2). d x 1 d t = 10 50 x 1 − 10 25 x 2 = 0.2 x 1 − 0.4 x 2 Consider matrix form of the system of equation as A. A = [ − 0.2 0.4 0.2 − 0.4 ] The characteristics equation of the coefficient matrix is, [ − 0.2 − λ 0.4 0.2 − 0.4 − λ ] = 0 ( − 0.2 − λ ) ( − 0.4 − λ ) − 0.08 = 0 λ 2 + 0.6 λ=0 λ = 0 or λ = − 0.6 The Eigen vector of the coefficient matrix A is in the form as, [ − 0.2 − λ 0.4 0.2 − 0.4 − λ ] [ a b ] = [ 0 0 ] The associated Eigen vector with the matrix Ais v = [ a b ] T . Substitute 0 for λ in coefficient matrix, [ − 0.2 + 0 0.4 0.2 − 0.4 + 0 ] [ a 1 b 1 ] = [ 0 0 ] [ 0.2 0.4 0.2 − 0.2 ] [ a 1 b 1 ] = [ 0 0 ] The equation can be obtained as, 0.2 a 1 + 0.4 b 1 = 0 0.2 a 1 − 0.4 b 1 = 0 After solving the equation the value of a 1 and b 1 can be obtained as, a 1 = 10 b 1 = 5 Therefore, the associated Eigen vector is shown below. v 1 = [ 10 5 ] Substitute − 0.6 for λ in coefficient matrix, [ − 0

5th Edition

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 29P

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Program Description:Purpose of problem is to find the solution of differential equation dx1dt=−k1x1+k2x2 and dx2dt=k1x1−k2x2 and also draw a graph of x1(t) and y1(t) .

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Chapter 5.2, Problem 28P

Chapter 5.2, Problem 30P

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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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Program Description: Purpose of problem is to find 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 y 1 ( t ) .

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

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Program Description:Purpose of problem is to find the solution of differential equation dx1dt=−k1x1+k2x2 and dx2dt=k1x1−k2x2 and also draw a graph of x1(t) and y1(t) .

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Chapter 5 Solutions