Program Explanation Given information: The differential function is x ″ − 4 x = 3 t . The initial condition given as x ( 0 ) = 0 and x ′ ( 0 ) = 0 . Explanation: The given differential function is x ″ − 4 x = 3 t . Apply Laplace transform to the function x ″ − 4 x = 3 t . L { x ″ − 4 x } = L { 3 t } L { x ″ } − L { 4 x } = L { ( 3 t ) } ( s 2 x ( s ) − x ( 0 ) s − x ′ ( 0 ) ) − 4 x ( s ) = 3 s 2 Apply initial condition x ( 0 ) = 0 and x ′ ( 0 ) = 0 in above equation. ( s 2 x ( s ) − ( 0 ) s − ( 0 ) ) − 4 x ( s ) = 3 s 2 x ( s ) = 3 s 2 x ( s ) = 3 s 2 ( s 2 − 4 ) Apply partial fraction method to solve the above expression, 3 s 2 ( s 2 − 4 ) = A s 2 + B ( s 2 − 4 ) ......

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ISBN: 9780100802452

Author: Calvis

Publisher: YUZU

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Chapter 7.3, Problem 29P

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Program Description: Purpose of problem is to solve the initial value problem x″−4x=3t by using Laplace transformation.

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Chapter 7.3, Problem 28P

Chapter 7.3, Problem 30P

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Chapter 7 Solutions

EBK DIFFERENTIAL EQUATIONS

Ch. 7.1 - Apply the definition in (1) to find directly tile...Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - Prob. 4P Ch. 7.1 - Prob. 5P Ch. 7.1 - Prob. 6P Ch. 7.1 - Prob. 7P Ch. 7.1 - Prob. 8P Ch. 7.1 - Prob. 9P Ch. 7.1 - Prob. 10P

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Program Description: Purpose of problem is to solve the initial value problem x ″ − 4 x = 3 t by using Laplace transformation.

Solution Summary: The author explains how to solve the initial value problem x′′-4x=3t by using Laplace transformation.

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Program Description: Purpose of problem is to solve the initial value problem x″−4x=3t by using Laplace transformation.

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Chapter 7 Solutions