Explanation Theorem used: Theorem 3.6.1 : Consider the non homogeneous second-order linear differential equation, y ″ + p ( t ) y ′ + q ( t ) y = g ( t ) If the functions p , q and g are continuous on an open interval I , and if the functions y 1 and y 2 form a fundamental set of solutions of the corresponding homogeneous equation y ″ + p ( t ) y ′ + q ( t ) y = 0 Then a particular solution of the given equation is, Y ( t ) = − y 1 ( t ) ∫ t 0 t y 2 ( s ) g ( s ) W [ y 1 , y 2 ] ( s ) d s + y 2 ( t ) ∫ t 0 t y 1 ( s ) g ( s ) W [ y 1 , y 2 ] ( s ) d s where t 0 is any conveniently chosen point in I . Then, the general solution is, y = c 1 y 1 ( t ) + c 2 y 2 ( t ) + Y ( t ) Given initial value problem is, L [ y ] = g ( t ) , y ( t 0 ) = 0 , y ′ ( t 0 ) = 0 According to the above mentioned theorem, Y ( t ) = − y 1 ( t ) ∫ t 0 t y 2 ( s ) g ( s ) W [ y 1 , y 2 ] ( s ) d s + y 2 ( t ) ∫ t 0 t y 1 ( s ) g ( s ) W [ y 1 , y 2 ] ( s ) d s = − y 1 ( t ) ∫ t 0 t y 2 ( s ) g ( s ) y 1 ( s ) y 2 ′ ( s ) − y 1 ′ ( s ) y 2 ( s ) d s + y 2 ( t ) ∫ t 0 t y 1 ( s ) g ( s ) y 1 ( s ) y 2 ′ ( s ) − y 1 ′ ( s ) y 2 ( s ) d s = ∫ t 0 t − y 1 ( t ) y 2 ( s ) g ( s ) y 1 ( s ) y 2 ′ ( s ) − y 1 ′ ( s ) y 2 ( s ) d s + ∫ t 0 t y 1 ( s ) y 2 ( t ) g ( s ) y 1 ( s ) y 2 ′ ( s ) −

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Question

Chapter 3.6, Problem 16P

To determine

To show: The expression for Y(t)=∫t0ty1(s)y2(t)−y1(t)y2(s)y1(s)y2′(s)−y1′(s)y2(s)g(s)ds and then show that, Y(t) is a solution of the given initial value problem.

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Chapter 3.6, Problem 15P

Chapter 3.6, Problem 17P

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Chapter 3 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

Ch. 3.1 - Prob. 1P Ch. 3.1 - Prob. 2P Ch. 3.1 - Prob. 3P Ch. 3.1 - Prob. 4P Ch. 3.1 - Prob. 5P Ch. 3.1 - Prob. 6P Ch. 3.1 - Prob. 7P Ch. 3.1 - Prob. 8P Ch. 3.1 - Prob. 9P Ch. 3.1 - Prob. 10P

Ch. 3.1 - Prob. 11P Ch. 3.1 - Prob. 12P Ch. 3.1 - Prob. 13P Ch. 3.1 - Prob. 14P Ch. 3.1 - Prob. 15P Ch. 3.1 - Prob. 16P Ch. 3.1 - Prob. 17P Ch. 3.1 - Prob. 18P Ch. 3.1 - Prob. 20P Ch. 3.2 - Prob. 1P Ch. 3.2 - Prob. 2P Ch. 3.2 - Prob. 3P Ch. 3.2 - Prob. 4P Ch. 3.2 - Prob. 5P Ch. 3.2 - Prob. 6P Ch. 3.2 - Prob. 7P Ch. 3.2 - Prob. 8P Ch. 3.2 - Prob. 9P Ch. 3.2 - Prob. 10P Ch. 3.2 - Prob. 11P Ch. 3.2 - Prob. 12P Ch. 3.2 - Prob. 13P Ch. 3.2 - Prob. 14P Ch. 3.2 - Prob. 15P Ch. 3.2 - Prob. 16P Ch. 3.2 - Prob. 17P Ch. 3.2 - Prob. 18P Ch. 3.2 - Prob. 19P Ch. 3.2 - Prob. 20P Ch. 3.2 - Prob. 21P Ch. 3.2 - Prob. 22P Ch. 3.2 - Prob. 23P Ch. 3.2 - Prob. 24P Ch. 3.2 - Prob. 25P Ch. 3.2 - Prob. 26P Ch. 3.2 - Prob. 27P Ch. 3.2 - Prob. 28P Ch. 3.2 - Prob. 29P Ch. 3.2 - Prob. 30P Ch. 3.2 - Prob. 31P Ch. 3.2 - Prob. 32P Ch. 3.2 - Prob. 33P Ch. 3.2 - Prob. 34P Ch. 3.2 - Prob. 35P Ch. 3.2 - Prob. 36P Ch. 3.2 - Prob. 37P Ch. 3.2 - Prob. 38P Ch. 3.3 - Prob. 1P Ch. 3.3 - Prob. 2P Ch. 3.3 - Prob. 3P Ch. 3.3 - Prob. 4P Ch. 3.3 - Prob. 5P Ch. 3.3 - Prob. 6P Ch. 3.3 - Prob. 7P Ch. 3.3 - Prob. 8P Ch. 3.3 - Prob. 9P Ch. 3.3 - Prob. 10P Ch. 3.3 - Prob. 11P Ch. 3.3 - Prob. 12P Ch. 3.3 - Prob. 13P Ch. 3.3 - Prob. 14P Ch. 3.3 - Prob. 15P Ch. 3.3 - Prob. 16P Ch. 3.3 - Prob. 17P Ch. 3.3 - Prob. 18P Ch. 3.3 - Prob. 19P Ch. 3.3 - Prob. 20P Ch. 3.3 - Prob. 21P Ch. 3.3 - Prob. 22P Ch. 3.3 - Prob. 23P Ch. 3.3 - Prob. 24P Ch. 3.3 - Prob. 25P Ch. 3.3 - Prob. 26P Ch. 3.3 - Prob. 27P Ch. 3.3 - Prob. 28P Ch. 3.3 - Prob. 29P Ch. 3.3 - Prob. 30P Ch. 3.3 - Prob. 31P Ch. 3.3 - Prob. 32P Ch. 3.3 - Prob. 33P Ch. 3.3 - Prob. 34P Ch. 3.3 - Prob. 35P Ch. 3.3 - Prob. 36P Ch. 3.4 - Prob. 1P Ch. 3.4 - Prob. 2P Ch. 3.4 - Prob. 3P Ch. 3.4 - Prob. 4P Ch. 3.4 - Prob. 5P Ch. 3.4 - Prob. 6P Ch. 3.4 - Prob. 7P Ch. 3.4 - Prob. 8P Ch. 3.4 - Prob. 9P Ch. 3.4 - Prob. 10P Ch. 3.4 - Prob. 11P Ch. 3.4 - Prob. 12P Ch. 3.4 - Prob. 13P Ch. 3.4 - Prob. 14P Ch. 3.4 - Prob. 15P Ch. 3.4 - Prob. 16P Ch. 3.4 - Prob. 17P Ch. 3.4 - Prob. 18P Ch. 3.4 - Prob. 19P Ch. 3.4 - Prob. 20P Ch. 3.4 - Prob. 21P Ch. 3.4 - Prob. 22P Ch. 3.4 - Prob. 23P Ch. 3.4 - Prob. 24P Ch. 3.4 - Prob. 25P Ch. 3.4 - Prob. 26P Ch. 3.4 - Prob. 27P Ch. 3.4 - Prob. 28P Ch. 3.4 - Prob. 29P Ch. 3.4 - Prob. 30P Ch. 3.4 - Prob. 31P Ch. 3.4 - Prob. 32P Ch. 3.4 - Prob. 33P Ch. 3.4 - Prob. 34P Ch. 3.5 - Prob. 1P Ch. 3.5 - Prob. 2P Ch. 3.5 - Prob. 3P Ch. 3.5 - Prob. 4P Ch. 3.5 - Prob. 5P Ch. 3.5 - Prob. 6P Ch. 3.5 - Prob. 7P Ch. 3.5 - Prob. 8P Ch. 3.5 - Prob. 9P Ch. 3.5 - Prob. 10P Ch. 3.5 - Prob. 11P Ch. 3.5 - Prob. 12P Ch. 3.5 - Prob. 13P Ch. 3.5 - Prob. 14P Ch. 3.5 - Prob. 15P Ch. 3.5 - Prob. 22P Ch. 3.5 - Prob. 23P Ch. 3.5 - Prob. 24P Ch. 3.5 - Prob. 25P Ch. 3.5 - Prob. 26P Ch. 3.5 - Prob. 27P Ch. 3.5 - Prob. 28P Ch. 3.5 - Prob. 29P Ch. 3.5 - Prob. 30P Ch. 3.6 - Prob. 1P Ch. 3.6 - Prob. 2P Ch. 3.6 - Prob. 3P Ch. 3.6 - Prob. 4P Ch. 3.6 - Prob. 5P Ch. 3.6 - Prob. 6P Ch. 3.6 - Prob. 7P Ch. 3.6 - Prob. 8P Ch. 3.6 - Prob. 9P Ch. 3.6 - Prob. 10P Ch. 3.6 - Prob. 11P Ch. 3.6 - Prob. 12P Ch. 3.6 - Prob. 13P Ch. 3.6 - Prob. 14P Ch. 3.6 - Prob. 15P Ch. 3.6 - Prob. 16P Ch. 3.6 - Prob. 17P Ch. 3.6 - Prob. 18P Ch. 3.6 - Prob. 19P Ch. 3.6 - Prob. 20P Ch. 3.6 - Prob. 21P Ch. 3.6 - Prob. 22P Ch. 3.6 - Prob. 23P Ch. 3.6 - Prob. 24P Ch. 3.6 - Prob. 25P Ch. 3.6 - Prob. 26P Ch. 3.7 - Prob. 1P Ch. 3.7 - Prob. 2P Ch. 3.7 - Prob. 3P Ch. 3.7 - Prob. 4P Ch. 3.7 - Prob. 5P Ch. 3.7 - Prob. 6P Ch. 3.7 - Prob. 7P Ch. 3.7 - Prob. 8P Ch. 3.7 - Prob. 9P Ch. 3.7 - Prob. 10P Ch. 3.7 - Prob. 11P Ch. 3.7 - Prob. 12P Ch. 3.7 - Prob. 13P Ch. 3.7 - Prob. 14P Ch. 3.7 - Prob. 15P Ch. 3.7 - Prob. 16P Ch. 3.7 - Prob. 17P Ch. 3.7 - Prob. 18P Ch. 3.7 - Prob. 19P Ch. 3.7 - Prob. 20P Ch. 3.7 - Prob. 21P Ch. 3.7 - Prob. 22P Ch. 3.7 - Prob. 23P Ch. 3.7 - Prob. 24P Ch. 3.8 - Prob. 1P Ch. 3.8 - Prob. 2P Ch. 3.8 - Prob. 3P Ch. 3.8 - Prob. 4P Ch. 3.8 - Prob. 5P Ch. 3.8 - Prob. 6P Ch. 3.8 - Prob. 7P Ch. 3.8 - Prob. 9P Ch. 3.8 - Prob. 11P Ch. 3.8 - Prob. 14P

Knowledge Booster

The expression for Y ( t ) = ∫ t 0 t y 1 ( s ) y 2 ( t ) − y 1 ( t ) y 2 ( s ) y 1 ( s ) y 2 ′ ( s ) − y 1 ′ ( s ) y 2 ( s ) g ( s ) d s and then show that, Y ( t ) is a solution of the given initial value problem.

To show: The expression for Y(t)=∫t0ty1(s)y2(t)−y1(t)y2(s)y1(s)y2′(s)−y1′(s)y2(s)g(s)ds and then show that, Y(t) is a solution of the given initial value problem.

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Chapter 3 Solutions