Explanation It is given as f ( t + T ) = f ( t ) . Apply Laplace transform on both sides: L { f ( t + T ) } = L { f ( t ) } ∫ 0 ∞ e − s t f ( t + T ) d t = ∫ 0 ∞ e − s t f ( t ) d t Assume as follows: u = t + T , then d u = d t when t = ∞ , u = ∞ and t = 0 , u = T Then, ∫ T ∞ e − s ( u − T ) f ( u ) d u = ∫ 0 ∞ e − s t f ( t ) d t ∫ T ∞ e − s u e s T f ( u ) d u = ∫ 0 T e − s t f ( t ) d t + ∫ T ∞ e − s t f ( t ) d t e s T ∫ T ∞ e − s u f ( u ) d u − ∫ T ∞ e − s t f ( t ) d t = ∫ 0 T e − s t f ( t ) d t ( e s T − 1 ) ∫ T ∞ e − s u f ( u ) d u = ∫ 0 T e − s

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Question

Chapter 6.3, Problem 24P

To determine

To show: The formula L{f(t)}=∫0Te−stf(t)dt1−e−sT holds.

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Chapter 6.3, Problem 23P

Chapter 6.3, Problem 25P

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Chapter 6 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

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