Explanation Given: The differential equation is, d x d t + 3 x + d y d t = 1 d x d t − x + d y d t − y = e t The initial values are x ( 0 ) = 0 , y ( 0 ) = 0 . Concept: The Linearity property of Laplace transformation is 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 + 3 x + d y d t } = L { 1 } L { d x d t − x + d y d t − y } = L { e t } Use the linearity property of Laplace transformation. L { d x d t } + 3 L { x } + L { d y d t } = L { 1 } L { d x d t } − L { x } + L { d y d t } − L { y } = L { e t } Use formula for derivatives, L { x ′ } | t = 0 = s L { x } − f ( 0 ) = s L { x } L { y ′ } | t = 0 = s L { y } − f ( 0 ) = s L { y } − 1 Substitute the Laplace transformation of the derivatives in linearity property. ( s + 3 ) L { x } + 5 L { y } = 1 s ( s − 1 ) L { x } + ( s − 1 ) L { y } = 1 s − 1 Simplify further and let L { x } = X ( s ) and L { y } = Y ( s ) . ( s + 3 ) X { s } + 5 Y { s } = 1 s ( s − 1 ) X { s } + ( s − 1 ) Y { s } = 1 s − 1 Solve the above equations by elimination method. X ( s ) = 1 − 2 s 3 s ( s − 1 ) 2 Y ( s ) = 5 s − 1 3 s ( s − 1 ) 2 Write X ( s ) as a sum of partial fractions. X ( s ) = 1 − 2 s 3 s ( s − 1 ) 2 = 1 3 { A s + B ( s − 1 ) + C ( s − 1 ) 2 } (2) Further solve the above expression as, 1 − 2 s 3 s ( s − 1 ) 2 = A ( s − 1 ) 2 + B s ( s − 1 ) + C s 3 s ( s − 1 ) 2 Compare the coefficients of the above expression

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 7.6, Problem 4E

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