Explanation Given: The function is e 2 t ( t − 1 ) 2 . Theorem used: First translation theorem, states that if L { f ( t ) } = F ( s ) and a is any real number, then L { e a t f ( t ) } = F ( s − a ) , where L { f ( t ) } denotes the Laplace transform of the function f ( t ) . Calculation: The Laplace transform of the function e 2 t ( t − 1 ) 2 is calculated as follows: L { e 2 t ( t − 1 ) 2 } = L { e 2 t ( t 2 − 2 t + 1 ) } = L { e 2 t t 2 − 2 e 2 t t + e 2 t } The linearity property of the Laplace transform is stated as follows: L { g 1 ( t ) + g 2 ( t ) + ⋯ } = L { g 1 ( t ) } + L { g 2 ( t ) } + ⋯ . Use the linearity property to calculate the Laplace transform of L { e 2 t t 2 − 2 e 2 t t + e 2 t } . L { e 2 t t 2 − 2 e 2 t t + e 2 t } = L { t 2 e 2 t } + L { − 2 t e 2 t } + L { e 2 t } = L { t 2 e 2 t } − 2 L { t e 2 t } + L { e 2 t } Use the result L { t n } = n ! s n + 1 and first translation theorem, L { e a t f ( t ) } = L { f ( t ) } | s → s − a . Here, s → s − a means to replace the symbol s by s − a wherever it appears

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Chapter 4.3, Problem 6E

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The Laplace transform of the function e2t(t−1)2.

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The Laplace transform of the function e 2 t ( t − 1 ) 2 .

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