Explanation Formula used: The solution of differential equation is y × I . F = ∫ Q × I . F d x + c , where I . F = e ∫ P d x . Calculation: The given differential equation y ′ + 2 y = g ( t ) with initial value condition y ( 0 ) = 0 . Where g ( t ) = 1 for 0 ≤ t < 1 and 0 for t > 1 . The value of I F is calculated as: I . F = e ∫ P d t = e ∫ 2 d t = e 2 t Substitute the value of integrating factor in the equation y × I . F = ∫ Q × I . F d x + c for 0 ≤ t < 1 . y × e 2 t = ∫ 1 × e 2 t d x y × e 2 t = ∫ e 2 t d x y × e 2 t = 1 2 e 2 t + c y = 1 2 + c e − 2 x Substitute the initial condition in above equation. y ( 0 ) = 1 2 + c e − 2 × 0 0 = 1 2 + c c = − 1 2 So, the value of solution of differential equation y = 1 2 + c e − 2 t . y = 1 2 − 1 2 e − 2 t y = 1 2 ( 1 − e − 2 t ) Substitute the value of integrating factor in the equation y × I . F = ∫ Q × I

Elementary Differential Equations and Boundary Value Problems, 11e WileyPLUS Registration Card + Loose-leaf Print Companion

11th Edition

ISBN: 9781119336617

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley (WileyPLUS Products)

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Question

Chapter 2.4, Problem 26P

To determine

To solve: The initial value problem of function y′+2y=g(t), when y(0)=0.

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Chapter 2.4, Problem 25P

Chapter 2.4, Problem 27P

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Chapter 2 Solutions

Elementary Differential Equations and Boundary Value Problems, 11e WileyPLUS Registration Card + Loose-leaf Print Companion

Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.1 - Prob. 10P

Ch. 2.1 - Prob. 11P Ch. 2.1 - Prob. 12P Ch. 2.1 - Prob. 13P Ch. 2.1 - Prob. 14P Ch. 2.1 - Prob. 15P Ch. 2.1 - Prob. 16P Ch. 2.1 - Prob. 17P Ch. 2.1 - Prob. 18P Ch. 2.1 - Prob. 19P Ch. 2.1 - Prob. 20P Ch. 2.1 - Prob. 21P Ch. 2.1 - Prob. 22P Ch. 2.1 - Prob. 23P Ch. 2.1 - Prob. 24P Ch. 2.1 - Prob. 25P Ch. 2.1 - Prob. 26P Ch. 2.1 - Prob. 27P Ch. 2.1 - Variation of Parameters. Consider the following...Ch. 2.1 - Prob. 29P Ch. 2.1 - In each of Problems 29 and 30, use the method of...Ch. 2.2 - Prob. 1P Ch. 2.2 - Prob. 2P Ch. 2.2 - Prob. 3P Ch. 2.2 - Prob. 4P Ch. 2.2 - Prob. 5P Ch. 2.2 - Prob. 6P Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - Prob. 9P Ch. 2.2 - Prob. 10P Ch. 2.2 - Prob. 11P Ch. 2.2 - Prob. 12P Ch. 2.2 - Prob. 13P Ch. 2.2 - Prob. 14P Ch. 2.2 - Prob. 15P Ch. 2.2 - Prob. 16P Ch. 2.2 - Prob. 17P Ch. 2.2 - Prob. 18P Ch. 2.2 - Prob. 19P Ch. 2.2 - Prob. 20P Ch. 2.2 - Prob. 21P Ch. 2.2 - Prob. 22P Ch. 2.2 - Prob. 23P Ch. 2.2 - Prob. 24P Ch. 2.2 - Prob. 25P Ch. 2.2 - Prob. 26P Ch. 2.2 - Prob. 27P Ch. 2.2 - Prob. 28P Ch. 2.2 - Prob. 29P Ch. 2.2 - Prob. 30P Ch. 2.2 - Prob. 31P Ch. 2.3 - Prob. 1P Ch. 2.3 - Prob. 2P Ch. 2.3 - Prob. 3P Ch. 2.3 - Prob. 4P Ch. 2.3 - Prob. 5P Ch. 2.3 - Prob. 6P Ch. 2.3 - Prob. 7P Ch. 2.3 - Prob. 8P Ch. 2.3 - Prob. 9P Ch. 2.3 - Prob. 10P Ch. 2.3 - Prob. 11P Ch. 2.3 - Prob. 12P Ch. 2.3 - Prob. 13P Ch. 2.3 - Prob. 14P Ch. 2.3 - Prob. 15P Ch. 2.3 - Prob. 16P Ch. 2.3 - Assume that the conditions are as in Problem 16...Ch. 2.3 - Prob. 18P Ch. 2.3 - Prob. 19P Ch. 2.3 - Prob. 20P Ch. 2.3 - Prob. 21P Ch. 2.3 - Prob. 22P Ch. 2.3 - Prob. 23P Ch. 2.3 - Prob. 24P Ch. 2.4 - In each of Problems 1 through 6, determine...Ch. 2.4 - Prob. 2P Ch. 2.4 - Prob. 3P Ch. 2.4 - Prob. 4P Ch. 2.4 - Prob. 5P Ch. 2.4 - Prob. 6P Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - Prob. 9P Ch. 2.4 - Prob. 10P Ch. 2.4 - Prob. 11P Ch. 2.4 - Prob. 12P Ch. 2.4 - Prob. 13P Ch. 2.4 - Prob. 14P Ch. 2.4 - Prob. 15P Ch. 2.4 - Prob. 16P Ch. 2.4 - Prob. 17P Ch. 2.4 - Prob. 18P Ch. 2.4 - Prob. 19P Ch. 2.4 - Prob. 20P Ch. 2.4 - Prob. 21P Ch. 2.4 - Prob. 22P Ch. 2.4 - Prob. 23P Ch. 2.4 - Prob. 24P Ch. 2.4 - Prob. 25P Ch. 2.4 - Prob. 26P Ch. 2.4 - Prob. 27P Ch. 2.5 - Prob. 1P Ch. 2.5 - Prob. 2P Ch. 2.5 - Prob. 3P Ch. 2.5 - Prob. 4P Ch. 2.5 - Prob. 5P Ch. 2.5 - Prob. 6P Ch. 2.5 - Prob. 7P Ch. 2.5 - Prob. 8P Ch. 2.5 - Prob. 9P Ch. 2.5 - Prob. 10P Ch. 2.5 - Prob. 11P Ch. 2.5 - Prob. 12P Ch. 2.5 - Prob. 13P Ch. 2.5 - Prob. 14P Ch. 2.5 - Prob. 15P Ch. 2.5 - Prob. 16P Ch. 2.5 - Prob. 17P Ch. 2.5 - Prob. 18P Ch. 2.5 - Prob. 19P Ch. 2.5 - Prob. 20P Ch. 2.5 - Prob. 21P Ch. 2.5 - Prob. 22P Ch. 2.5 - Prob. 23P Ch. 2.5 - Prob. 24P Ch. 2.5 - Prob. 25P Ch. 2.5 - Prob. 26P Ch. 2.5 - Prob. 27P Ch. 2.6 - Prob. 1P Ch. 2.6 - Prob. 2P Ch. 2.6 - Prob. 3P Ch. 2.6 - Prob. 4P Ch. 2.6 - Prob. 5P Ch. 2.6 - Prob. 6P Ch. 2.6 - Prob. 7P Ch. 2.6 - Prob. 8P Ch. 2.6 - Prob. 9P Ch. 2.6 - Prob. 10P Ch. 2.6 - Prob. 11P Ch. 2.6 - Prob. 12P Ch. 2.6 - Prob. 13P Ch. 2.6 - Prob. 14P Ch. 2.6 - Prob. 15P Ch. 2.6 - Prob. 16P Ch. 2.6 - Prob. 17P Ch. 2.6 - Prob. 18P Ch. 2.6 - Prob. 19P Ch. 2.6 - Prob. 20P Ch. 2.6 - Prob. 21P Ch. 2.6 - Prob. 22P Ch. 2.7 - Prob. 1P Ch. 2.7 - Prob. 2P Ch. 2.7 - Prob. 3P Ch. 2.7 - Prob. 4P Ch. 2.7 - Prob. 5P Ch. 2.7 - Prob. 6P Ch. 2.7 - Prob. 7P Ch. 2.7 - Prob. 8P Ch. 2.7 - Prob. 9P Ch. 2.7 - Prob. 10P Ch. 2.7 - Prob. 11P Ch. 2.7 - Prob. 12P Ch. 2.7 - Prob. 14P Ch. 2.7 - Prob. 15P Ch. 2.7 - Prob. 16P Ch. 2.7 - Prob. 17P Ch. 2.8 - Prob. 1P Ch. 2.8 - Prob. 2P Ch. 2.8 - Prob. 3P Ch. 2.8 - Prob. 4P Ch. 2.8 - Prob. 5P Ch. 2.8 - Prob. 6P Ch. 2.8 - Prob. 7P Ch. 2.8 - Prob. 8P Ch. 2.8 - Prob. 9P Ch. 2.8 - Prob. 10P Ch. 2.8 - Prob. 11P Ch. 2.8 - Prob. 12P Ch. 2.8 - Prob. 13P Ch. 2.8 - Prob. 14P Ch. 2.8 - Prob. 15P Ch. 2.8 - Prob. 16P Ch. 2.8 - Prob. 17P Ch. 2.8 - Prob. 18P Ch. 2.9 - Prob. 1P Ch. 2.9 - Prob. 2P Ch. 2.9 - Prob. 3P Ch. 2.9 - Prob. 4P Ch. 2.9 - Prob. 5P Ch. 2.9 - Prob. 6P Ch. 2.9 - Prob. 7P Ch. 2.9 - Prob. 8P Ch. 2.9 - Prob. 9P Ch. 2.9 - Prob. 10P Ch. 2 - Prob. 1MP Ch. 2 - Prob. 2MP Ch. 2 - Prob. 3MP Ch. 2 - Prob. 4MP Ch. 2 - Prob. 5MP Ch. 2 - Prob. 6MP Ch. 2 - Prob. 7MP Ch. 2 - Prob. 8MP Ch. 2 - Prob. 9MP Ch. 2 - Prob. 10MP Ch. 2 - Prob. 11MP Ch. 2 - Prob. 12MP Ch. 2 - Prob. 13MP Ch. 2 - Prob. 14MP Ch. 2 - Prob. 15MP Ch. 2 - Prob. 16MP Ch. 2 - Prob. 17MP Ch. 2 - Prob. 18MP Ch. 2 - Prob. 19MP Ch. 2 - Prob. 20MP Ch. 2 - Prob. 21MP Ch. 2 - Prob. 22MP Ch. 2 - Prob. 23MP Ch. 2 - Prob. 24MP Ch. 2 - Prob. 25MP Ch. 2 - Prob. 28MP Ch. 2 - Prob. 29MP Ch. 2 - Prob. 31MP Ch. 2 - Prob. 32MP Ch. 2 - Prob. 33MP Ch. 2 - Prob. 34MP Ch. 2 - Prob. 35MP Ch. 2 - Prob. 36MP Ch. 2 - Prob. 37MP

Knowledge Booster

The initial value problem of function y ′ + 2 y = g ( t ) , when y ( 0 ) = 0 .

To solve: The initial value problem of function y′+2y=g(t), when y(0)=0.

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Chapter 2 Solutions