To determine To show: The expression e t 2 does not satisfy the condition | f ( t ) ≤ M e k t | . Explanation Definition used: A function f ( t ) has Laplace transform if for all t ≥ 0 and some constants M and k it satisfies the growth restriction | f ( t ) | ≤ M e k t Calculation: From the above definition, t ≥ 0 . That is, e t 2 > 0 . It remains to prove that, there exist t such that e t 2 > M e k t . Take logarithms on both sides of the inequality e t 2 > M e k t as, log ( e t 2 ) > log ( M e k t ) t 2 > log ( M ) + log ( e k t ) t 2 >

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 6.1, Problem 20P

To determine

To show: The expression et2 does not satisfy the condition |f(t)≤Mekt|.

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Chapter 6.1, Problem 19P

Chapter 6.1, Problem 21P

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