Explanation Given: The provided differential equation is, d x d t + x − d y d t + y = 0 d x d t + d y d t + 2 y = 0 Provided initial values are x ( 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 ) } 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 x d t + x − d y d t + y } = L { 0 } L { d x d t + d y d t + 2 y } = L { 0 } Use the linearity property of Laplace transformation. L { d x d t } + L { x } − L { d y d t } + L { y } = L { 0 } ...... (1) L { d x d t } + L { d y d t } + 2 L { y } = L { 0 } Use formula for Laplace of derivatives, L { x ' } | t = 0 = s L { x } − x ( 0 ) = s L { x } And L { y ′ } | t = 0 = s L { y } − y ( 0 ) = s L { y } − 1 Substitute the Laplace transformation of the derivatives in equation (1) s X − x ( 0 ) + X − s Y + y ( 0 ) + Y = 0 s X + X − s Y + Y + 1 = 0 Simplify further, ( ( s + 1 ) ( s + 2 ) + ( s 2 − s ) ) X = − s − 2 + s − 1 ( 2 s 2 + 2 s + 2 ) X = − 3 And ( s + 2 ) Y = 3 s 2 ( s 2 + s + 1 ) + 1 Solve the above equations by eliminate, X ( s ) = − 3 2 1 ( s 2 + s + 1 ) Y ( s ) = 1 2 2 s + 1 ( s 2 + s + 1 ) Write X ( s ) in this form

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.6, Problem 6E

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Chapter 4 Solutions

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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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Chapter 4 Solutions