Explanation Given: The provided step function is f ( t ) = { 1 , 0 ≤ t < 1 − 1 , t ≥ 1 Concept: The unit step function U ( t − a ) is defined as, U ( t − a ) = { 0 , 0 ≤ t < a 1 , t ≥ a The general piecewise-defined function of the type, f ( t ) = { g ( t ) , 0 ≤ t < a h ( t ) , t ≥ a Where, it is written as f ( t ) = g ( t ) − g ( t ) U ( t − a ) + h ( t ) U ( t − a ) . The second translation theorem is If F ( s ) = L { f ( t ) } and a > 0 then L { f ( t − a ) U ( t − a ) } = e − a s F ( s ) The Laplace transformation of constant L { a } = a s . Calculation: The provided function f ( t ) = { 1 , 0 ≤ t < 1 − 1 , t ≥ 1 . Compare above function with the general piecewise function, get values as, g ( t ) = 1 , h ( t ) = − 1 and a = 1 Put values of g ( t ) , h ( t ) and a in f ( t ) = g ( t ) − g ( t ) U ( t − a ) + h (

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.3, Problem 64E

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To solve: The given initial value problem using Laplace transforms.

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Chapter 4.3, Problem 63E

Chapter 4.3, Problem 65E

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Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The given initial value problem using Laplace transforms.

Solution Summary: The author explains the Laplace transforms of the given initial value problem.

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To solve: The given initial value problem using Laplace transforms.

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Chapter 4 Solutions