Program Explanation Given information: The convolution property states as, L − 1 { F ( s ) . G ( s ) } = f ( t ) * g ( t ) = ∫ 0 t f ( τ ) g ( t − τ ) d τ If Laplace transform defined as L { f ( t ) } = F ( s ) and L { g ( t ) } = G ( s ) . Calculation: Apply the Convolution theorem to solve the inverse Laplace expression L − 1 { 1 ( s − 1 ) s } as, L − 1 { 1 ( s − 1 ) s } = L − 1 { 1 ( s − 1 ) } * L − 1 { 1 s } = e t * t − 1 2 Γ ( 1 2 ) = e t * 1 t π = 1 π (

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ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 7.4, Problem 35P

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Program Description: Purpose of problem is to show that L−1{1(s−1)s}=2etπ∫0te−u2du=eterft .

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Chapter 7.4, Problem 34P

Chapter 7.4, Problem 36P

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Chapter 7 Solutions

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Ch. 7.1 - Apply the definition in (1) to find directly tile...Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - Prob. 4P Ch. 7.1 - Prob. 5P Ch. 7.1 - Prob. 6P Ch. 7.1 - Prob. 7P Ch. 7.1 - Prob. 8P Ch. 7.1 - Prob. 9P Ch. 7.1 - Prob. 10P

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Program Description: Purpose of problem is to show that L − 1 { 1 ( s − 1 ) s } = 2 e t π ∫ 0 t e − u 2 d u = e t erf t .

Solution Summary: The author explains how the Convolution theorem solves the inverse Laplace expression.

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Program Description: Purpose of problem is to show that L−1{1(s−1)s}=2etπ∫0te−u2du=eterft .

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Chapter 7 Solutions