Explanation Given: The given linear differential equation is y ″ + y = sin t , and given initial values are y ( 0 ) = 1 and y ′ ( 0 ) = − 1 . Formula used: 1. The Laplace transformation of a linear differential equation with constant coefficient is express as, a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] 2. The formula for the inverse Laplace transform is as follows, L − 1 { k s 2 + k 2 } = sin k t L − 1 { s s 2 + k 2 } = cos k t L − 1 { s 2 − k 2 ( s 2 + k 2 ) 2 } = t cos k t 3. The formula for the Laplace transform of sin k t is as follows, L { sin k t } = k s 2 + k 2 4. If L − 1 { F ( s ) } = f ( t ) and for any value of a , the property of inverse Laplace transform is, L − 1 { ( − 1 ) n d n d s n { F ( s ) } } = t n f ( t ) 5. The linearity property of the Laplace transformation is as below. If a , b be any constant and f , g be any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } Calculation: Use the above definition to write the Laplace transform of the given linear differential equation. s 2 L { y } − s y ( 0 ) − y ′ ( 0 ) + L { y } = L { sin t } Put the value y ( 0 ) = 1 and y ′ ( 0 ) = − 1 in above expression. ( s 2 + 1 ) L { y } − s + 1 = 1 s 2 + 1 ( s 2 + 1 ) L { y } = 1 s 2 + 1 + s − 1 ( s 2 + 1 ) L { y } = s 3 − s 2 + s ( s 2 + 1 ) L { y } = s 3 − s 2 + s ( s 2 + 1 ) 2 From the above expression, L { y } = s 3 − s 2 + s ( s 2 + 1 ) 2 . (1) Equation (1) is simplified using partial fraction as follows s 3 − s 2 + s ( s 2 + 1 ) 2 = a 1 s + a 0 s 2 + 1 + a 3 s + a 2 ( s 2 + 1 ) 2 From the above expression, s 3 − s 2 + s ( s 2 + 1 ) 2 = a 1 s + a 0 s 2 + 1 + a 3 s + a 2 ( s 2 + 1 ) 2 . (2) Multiply with ( s 2 + 1 ) 2 on both side of equation (2)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.4, Problem 12E

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To solve: The given initial-value problem using Laplace transform.

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Chapter 4.4, Problem 11E

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The given initial-value problem using Laplace transform.

Solution Summary: The author explains the Laplace transform of a linear differential equation with constant coefficient.

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To solve: The given initial-value problem using Laplace transform.

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