Explanation Given: The provided differential equation is, d 2 x d t 2 + d 2 y d t 2 = t 2 d 2 x d t 2 − d 2 y d t 2 = 4 t Provided initial values are x ( 0 ) = 8 , x ' ( 0 ) = − 0 y ( 0 ) = 0 , 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 + d 2 y d t 2 } = L { t 2 } L { d 2 x d t 2 − d 2 y d t 2 } = L { 4 t } Use the linearity property of Laplace transformation. L { d 2 x d t 2 } + L { d 2 y d t 2 } = L { t 2 } L { d 2 x d t 2 } − L { d 2 y d t 2 } = L 4 { t } Use formula for Laplace of derivatives, L { x ' } | t</

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Chapter 4.6, Problem 9E

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To Solve: The system of differential equation with the use of Laplace transforms.

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Chapter 4.6, Problem 8E

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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