Explanation The given function is e − s s ( s + 1 ) . From inverse form of Second Translation theorem, if f ( t ) = L − 1 { F ( s ) } and a > 0 , then L − 1 { e − a s F ( s ) } = f ( t − a ) U ( t − a ) . Here, F ( s ) = 1 s ( s + 1 ) and a = 1 . Therefore, L − 1 { F ( s ) } = L − 1 { 1 s ( s + 1 ) } By partial fraction it can be written as, 1 s ( s + 1 ) = A s + B s + 1 . Solving this gives A = 1 and B = − 1 . Thus, L − 1 { F ( s ) } = L − 1 { 1 s ( s + 1 ) } = L − 1 { 1 s } − L − 1 {

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.3, Problem 47E

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The inverse Laplace transformation of the given function.

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Chapter 4.3, Problem 46E

Chapter 4.3, Problem 48E

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Chapter 4 Solutions

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The inverse Laplace transformation of the given function.

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