Explanation Given: The function is, F ( s ) = e − 4 s ( s + 3 ) s 2 − 6 s + 13 . Approach: From First Shifting Theorem, Let L − 1 [ F ( s ) ] = f ( t ) . Then L − 1 [ F ( s − a ) ] = e a t f ( t ) . From Second Shifting Theorem, Let L − 1 [ F ( s ) ] = f ( t ) . Then L − 1 [ e − a s F ( s ) ] = u a ( t ) f ( t − a ) . Calculation: The function is, F ( s ) = e − 4 s ( s + 3 ) s 2 − 6 s + 13 Compare the above function with L − 1 [ e − a s F ( s ) ] . a = 4 , and F ( s ) = ( s + 3 ) s 2 − 6 s + 13 , so t must be shifted to ( t − 4 ) From first Shifting Theorem, the function is written as, L − 1 [ ( s + 3 ) s 2 − 6 s + 13 ] = L − 1 [ ( s − 3 ) ( s − 3 ) 2 + 4 + 6 ( 1 ( s − 3 ) 2 + 4 ) ] = e 3 t cos 2 t + 3 e 3 t sin 2 t From Second Shifting Theorem, the function is written as,

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 10.7, Problem 23P

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The inverse Laplace transform of F.

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