Explanation Given: The function is f ( t ) = e − 1 s . The equality is L − 1 { s − 1 / 2 e − 1 / s } ( t ) = 1 π t cos 2 t . Approach: 1) Taylor series of a function g ( t ) about the point t = a is given by, g ( t ) = g ( a ) + ( t − a ) g ′ ( a ) + ( t − a ) 2 2 ! g ″ ( a ) + ... 2) The Laplace transforms satisfies the property: i. L { t n } ( s ) = n ! s n + 1 . ii. L − 1 { 1 s ( n + 1 2 ) } ( t ) = 2 n t n − 1 2 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2 n − 1 ) π 3) The series expansion of cos x is given by cos x = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! Calculation: The series expansion of e x is given by, e s = 1 + s + s 2 2 ! + s 3 3 ! + s 4 4 ! + ... ... ( 1 ) Replace s by − 1 s in the Equation ( 1 ) . The following series is obtained: e − 1 s = 1 − 1 s + 1 2 ! s 2 − 1 3 ! s 3 + 1 4 ! s 4 + ... e − 1 s = ∑ n = 0 ∞ ( − 1 s ) n 1 n ! Multiply s − 1 / 2 e − 1 / s on both sides of the above series. s − 1 / 2 e − 1 / s = s − 1 / 2 ( ∑ n = 0 ∞ ( − 1 s ) n 1 n ! ) = ∑ n = 0 ∞ ( − 1 s ) n s − 1 / 2 1 n ! = ∑ n = 0 ∞ ( − 1 ) n s − n s − 1 / 2 1 n ! = ∑ n = 0 ∞ ( − 1 ) n 1 s ( n + 1 2 ) 1 n ! … ( 2 ) Take inverse Laplace transform in Equation ( 2 ) . L − 1 { s − 1 / 2 e − 1 / s } ( t ) = L − 1 { ∑ n = 0 ∞ ( − 1 ) n 1 s ( n + 1 2 ) 1 n ! } ( t ) = ∑ n = 0 ∞ ( − 1 ) n n ! L − 1 { 1 s ( n + 1 2 ) } ( t ) = ∑ n = 0 ∞ ( − 1 ) n n ! 2 n t n − 1 2 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2 n − 1 ) π = 1 π t ∑ n = 0 ∞ ( − 1 ) n n ! 2 n t n 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2 n − 1 )

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 7.7, Problem 23E

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The expansion for e−1s in powers of 1s and s−1/2e−1/s in terms of 1/sn+1/2 and show that L−1{s−1/2e−1/s}(t)=1πtcos2t.

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Chapter 7.7, Problem 22E

Chapter 7.7, Problem 24E

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The expansion for e − 1 s in powers of 1 s and s − 1 / 2 e − 1 / s in terms of 1 / s n + 1 / 2 and show that L − 1 { s − 1 / 2 e − 1 / s } ( t ) = 1 π t cos 2 t .

Solution Summary: The author explains that the expansion for e-1s in powers of

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The expansion for e−1s in powers of 1s and s−1/2e−1/s in terms of 1/sn+1/2 and show that L−1{s−1/2e−1/s}(t)=1πtcos2t.

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