To determine The Laplace transformation of the function f ( t ) = cosh b t . 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 ) = cosh b t . Compute the Laplace transformation of f ( t ) = cosh b t as follows. Substitute f ( t ) = cosh b t in F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t and evaluate further. F ( s ) = ∫ 0 ∞ e − s t cosh b t d t . Let F ( s ) = lim A → ∞ ∫ 0 A e − s t cosh b t d t (1) It is given that, cosh b t = e b t + e − b t 2 . So, the above equation will become as, F ( s ) = lim A → ∞ ∫ 0 A e − s t ( e b t + e − b t 2 ) d t = lim A → ∞ [ 1 2 ∫ 0 A e b t ⋅ e − s t d t + 1 2 ∫ 0 A e − b t ⋅ e − s t d t ] = lim A → ∞ [ 1 2 ∫ 0 A e ( b − s ) t d t + 1 2 ∫ 0 A e − ( b + s ) t d t ] Let F ( s ) = lim A → ∞ [ 1 2 ∫ 0 A e ( b − s ) t d t + 1 2 ∫ 0 A e − ( b + s ) t d t ] be equation (2). First integrate ∫ 0 A e ( b − s ) t d t with respect to t as follows. Let u = ( b − s ) t , then d u = ( b − s ) d t . Note that, for t = 0, u = 0 and for t = A , u = ( b − s ) A . That implies, ∫ 0 A e ( b − s ) t d t = ∫ 0 ( b − s ) A e u b − s d u = 1 b − s ∫ 0 ( b − s ) A e u d u = 1 b − s [ e u ] 0 ( b − s ) A = 1 b − s [ e ( b − s ) A − e 0 ] = 1 b − s [ e A ( b − s ) − 1 ] That is, ∫ 0 A e ( b − s ) t d t = 1 b − s [ e A ( b − s ) − 1 ]

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 6.1, Problem 7P

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The Laplace transformation of the function f(t)=coshbt.

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Chapter 6.1, Problem 6P

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Elementary Differential Equations

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