Program Explanation Given information: The system of differential equation is x ′ = 3 x + 4 y and y ′ = 3 x + 2 y + t 2 . The initial condition is x ( 0 ) = y ( 0 ) = 0 . Explanation: Consider matrix form of the system of differential equation as, x ′ = [ 3 4 3 2 ] x + [ 0 t 2 ] ...............(1) The homogeneous solution [ 3 4 3 2 ] can be obtained as, | A − λ I | = 0 [ 3 − λ 4 3 2 − λ ] = 0 ( 3 − λ ) ( 2 − λ ) − 12 = 0 λ 2 − 5 λ − 6 = 0 Further solve the above equation as, λ 2 − 5 λ − 6 = 0 ( λ − 6 ) ( λ + 1 ) = 0 λ = − 1 , 6 The corresponding Eigen vector of the equation is, v 1 = [ 1 − 1 ] and v 2 = [ 4 3 ] The complementary solution of the equation is, x c ( t ) = c 1 e − t + 4 c 2 e 6 t y c ( t ) = − c 1 e − t + 3 c 2 e 6 t Therefore, linear trial particular solution is in the form as, x p ( t ) = a 1 + b 1 t + c 1 t 2 y p ( t ) = a 2 + b 2 t + c 2 t 2 Solve equation (1) by substituting the valueof x p ( t ) , y p ( t ) [ b 1 + 2 c 1 b 2 + 2 c 2 ] = [ 3 4 3 2 ] [ a 1 + b 1 t + c 1 t 2 a 2 + b 2 t + c 2 t 2 ] + [ 0 t 2 ] [ b 1 + 2 c 1 b 2 + 2 c 2 ] = [ 3 ( a 1 + b 1 t + c 1 t 2 ) + 4 ( a 2 + b 2 t + c 2 t 2 ) 2 ( a 1 + b 1 t + c 1 t 2 ) + 2 ( a 2 + b 2 t + c 2 t 2 ) ] + [ 0 t 2 ] [ b 1 + 2 c 1 b 2 + 2 c 2 ] = [ 3 ( a 1 + b 1 t + c 1 t 2 ) + 4 ( a 2 + b 2 t + c 2 t 2 ) 2 ( a 1 + b 1 t + c 1 t 2 ) + 2 ( a 2 + b 2 t + c 2 t 2 ) + t 2 ] Equate the coefficient to obtain equations are, 3 a 1 + 4 a 2 = b 1 3 a 1 + 2 a 2 = b 2 3 b 1 + 4 b 2 = 2 c 1 3 b 1 + 2 b 2 = 2 c 2 <

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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 3P

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Program Description:Purpose of problem is to find a particular solution of the system x′=3x+4y and y′=3x+2y+t2 by using method of undetermined coefficient also apply initial condition to verify the solution.

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Chapter 5.7, Problem 2P

Chapter 5.7, Problem 4P

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Chapter 5 Solutions

EP DIFF.EQUAT.+BOUND.VALUE,...UPD.-ACC.

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 find a particular solution of the system x ′ = 3 x + 4 y and y ′ = 3 x + 2 y + t 2 by using method of undetermined coefficient also apply initial condition to verify the solution.

Solution Summary: The author explains that the problem is to find a particular solution of the system xprime by using method of undetermined coefficient.

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Program Description:Purpose of problem is to find a particular solution of the system x′=3x+4y and y′=3x+2y+t2 by using method of undetermined coefficient also apply initial condition to verify the solution.

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Chapter 5 Solutions