Explanation Formula used: The Laplace transformation of f : “The Laplace transform of f which is dented by L { f ( t ) } = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t ”. Calculation: The given function is f ( t ) = { t , 0 ≤ t < 1 2 − t , 1 ≤ t < 2 0 , 2 ≤ t < ∞ } . Compute the Laplace transformation of f ( t ) = { t , 0 ≤ t < 1 2 − t , 1 ≤ t < 2 0 , 2 ≤ t < ∞ } as follows. Substitute f ( t ) = t for 0 ≤ t < 1 , f ( t ) = 2 − t for 1 ≤ t < 2 , and f ( t ) = 0 for 2 ≤ t < ∞ in F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t and evaluate further. F ( s ) = ∫ 0 1 e − s t ( t ) d t + ∫ 1 2 e − s t ( 2 − t ) d t + ∫ 2 ∞ e − s t ( 0 ) d t F ( s ) = ∫ 0 1 t e − s t d t + ∫ 1 2 ( 2 − t ) e − s t d t Let F ( s ) = ∫ 0 1 t e − s t d t + ∫ 1 2 ( 2 − t ) e − s t d t be equation (1). First compute the value of ∫ 0 1 t e − s t d t as follows. Here, u = t , then d u = d t and d v = e − s t d t . That implies, ∫ d v = ∫ e − s t d t v = − 1 s ∫ e u d u [ let u = − s t , then d u = − s d t ] = − 1 s [ e u ] = − e − s t s The equation (1) becomes, ∫ 0 1 t e − s t d t = [ t ⋅ − e − s t s ] 0 1 − ∫ 0 1 − e − s t s d t = [ 1 ⋅ − e − s s − 0 ] + 1 s ∫ 0 1 e − s t d t = − e − s s + 1 s ∫ 0 1 − 1 s e w d w [ let w = − s t , then d w = − s d t ] = − e − s s − 1 s 2 [ e w ] 0 1 On further simplification, ∫ 0 1 t e − s t d t = − e − s s − 1 s 2 [ e − s t ] 0 1 [ ∵ w = − s t ] = − e − s s − 1 s 2 [ e − s − 1 ] = − e − s s − e − s s 2 + 1 s 2 = 1 − s e − s − e − s s 2 = 1 − ( s + 1 ) e − s s 2 Now compute the value of ∫ 1 2 ( 2 − t ) e − s t d t as follows

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 6.1, Problem 24P

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The Laplace transformation of the given function.

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Elementary Differential Equations

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