Explanation Given: The provided differential equation is, d 2 x d t 2 + x − y = 0 d 2 y d t 2 + y − x = 0 Provided initial values are x ( 0 ) = 0 , x ' ( 0 ) = − 2 y ( 0 ) = 0 , y ' ( 0 ) = 1 Concept: Linearity property of Laplace transformation is as below. L [ a f ( t ) + b g ( t ) ] = a L { f ( t ) } + b L { g ( t ) } ...... (1) 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 + x − y } = L { 0 } L { d 2 y d t 2 + y − x } = L { 0 } Use the linearity property of Laplace transformation. L { d 2 x d t 2 } + L { x } − L { y } = L { 0 } L { d 2 y d t 2 } + L { y } − L { x } = L { 0 } 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 + 1 ) L { x } − L { y } = − 2 ( s 2 + 1 ) L { y } − L { x } = 1 Simplify further, and let L { x } = X ( s ) and L { y } = Y ( s ) ( s 2 + 1 ) X { s } − Y { s } = − 2 ( s 2 + 1 ) Y { s } − X { s } = 1 Solve the above equations by eliminate, X ( s ) = − 2 s 2 − 1 s 2 ( s 2 + 2 ) Y ( s ) = s 2 − 1 s 2 ( s 2 + 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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The system of differential equation with the use of Laplace transforms.

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