Explanation Given: The provided differential equation is, d 2 x d t 2 + 3 d y d t + 3 y = 0 d 2 x d t 2 + 3 y = t e − t Provided initial values are x ( 0 ) = 0 , x ' ( 0 ) = 2 , y ( 0 ) = 0 Concept: Linearity property of Laplace transformation is as below. L [ a f ( t ) + b g ( t ) ] = a L { f ( t ) } + b L { g ( t ) } Where a and b are any constants and f and g be any functions of t . The formula for the transform of a derivative in Laplace transformation, L { f ( n ) ( t ) } = s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ' ( 0 ) ⋯ − f ( n − 1 ) ( 0 ) Where F ( s ) = L { f ( t ) } The formula for the inverse Laplace transform of e − a t as below. L − 1 { 1 s + a } = e − a t Calculation: Use the above definition to write the Laplace transform of the given differential equation. L { d 2 x d t 2 + 3 d y d t + 3 y } = L { 0 } L { d 2 x d t 2 + 3 y } = L { t e − t } Use the linearity property of Laplace transformation. L { d 2 x d t 2 } + 3 L { d y d t } + 3 L { y } = L { 0 } L { d 2 x d t 2 } + 3 L { y } = L { t e − t } Use formula for Laplace of derivatives, L { x ' } | t = 0 = s L { x } − f ( 0 ) = s L { x } And L { y ′ } | t = 0 = s L { y } − f ( 0 ) = s L { y } − 1 Substitute the Laplace transformation of the derivatives in linearity property, s 2 L { x } + 3 ( s + 1 ) L { y } = 0 s 2 L { x } + 3 L { y } = 1 ( s + 1 ) 2 Simplify further, and let L { x } = X ( s ) and L { y } = Y ( s ) s 2 X { s } + 3 ( s + 1 ) Y { s } = 0 s 2 X { s } + 3 Y { s } = 1 ( s + 1 ) 2 Solve the above equations by eliminate, X ( s ) = − 2 s + 1 s 3 ( s + 1 ) Y ( s ) = 1 3 ( 2 s 2 + 2 s + 1 ( s + 1 ) 2 ) Write X ( s ) as a sum of partial fractions

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 4.6, Problem 11E

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The system of differential equation with the use of Laplace transforms.

Solution Summary: The author explains the Laplace transformation of differential equation. The formula for the transform of a derivative is as follows.

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