Explanation Given: The linear differential equations are x ″ + y ″ = e 2 t and 2 x ′ + y ″ = − e 2 t . The initial values are x ( 0 ) = 0 , y ( 0 ) = 0 , x ′ ( 0 ) = 0 and y ′ ( 0 ) = 0 . Concept used: (1) The Laplace transformation of a linear differential equation with constant coefficient is expressed as follows: a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] (2) The linearity property of the Laplace transformation is stated as, if a , b are any constant and f , g are any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } Calculation: The differential equation is given below: x ″ + y ″ = e 2 t … … ( 1 ) 2 x ′ + y ″ = − e 2 t … … ( 2 ) Subtract equation ( 2 ) from equation ( 1 ) . x ″ − 2 x ′ = 2 e 2 t … … ( 3 ) Use the above definition (1) to write the Laplace transform of the equation ( 3 ) . s 2 L { x } − s x ( 0 ) − x ′ ( 0 ) − 2 [ s L { x } − x ( 0 ) ] = 2 L { e 2 t } Substitute the value of x ( 0 ) as 0 and x ′ ( 0 ) as 0 in above equation. ( s 2 − 2 s ) L { x } = 2 ⋅ 1 s − 2 L { x } = 2 s ( s − 2 ) 2 … … ( 4 ) Solve the equation ( 4 ) by partial fraction. 2 s ( s − 2 ) 2 = a 0 s + a 1 s − 2 + a 2 ( s − 2 ) 2 … … ( 5 ) Multiply s ( s − 2 ) 2 to equation ( 5 ) . 2 = a 0 ( s − 2 ) 2 + a 1 s ( s − 2 ) + a 2 s … … ( 6 ) Substitute 0 for s in equation ( 6 ) . 2 = a 0 ( 0 − 2 ) 2 + a 1 ⋅ 0 ⋅ ( 0 − 2 ) + a 2 ⋅ 0 2 = 4 a 0 a 0 = 1 2 ( Divide by 4 ) Substitute 2 for s in equation ( 6 ) . 2 = a 0 ( 2 − 2 ) 2 + a 1 ⋅ 2 ( 2 − 2 ) + a 2 ⋅ 2 2 = 2 a 2 a 2 = 1 ( Divide by 2 ) Substitute 1 4 for a 1 and 1 for a 2 in equation ( 6 ) . 2 = 1 2 ( s − 2 ) 2 + a 1 s ( s − 2 ) + 1 ⋅ s 2 = a 1 s 2 + 1 2 s 2 − 2 a 1 s − s + 2 2 = s 2 ( a 1 + 1 2 ) + s ( − 2 a 1 − 1 ) + 2 Compare the similar coefficient of s 2 on both sides of above equation. a 1 + 1 2 = 0 a 1 = − 1 2 Substitute 1 2 for a 0 , − 1 2 for a 1 and 1 for a 2 in equation ( 5 ) . 2 s ( s − 2 ) 2 = 1 2 s − 1 2 ( s − 2 ) + 1 ( s − 2 ) 2 Now equation ( 4 ) is written as follows: L { x } = 1 2 s − 1 2 ( s − 2 ) + 1 ( s − 2 ) 2 Take inverse Laplace on both sides of above equation. x = L − 1 { 1 2 s − 1 2 ( s − 2 ) + 1 ( s − 2 ) 2 } = 1 2 L − 1 { 1 s } − 1 2 L − 1 { 1 s − 2 } + L − 1 { 1 ( s − 2 ) 2 } = 1 2 − 1 2 e 2 t + t e 2 t Now add equation ( 1 ) and equation ( 2 ) . x ″ + 2 x ′ + 2 y ″ = 0 … … ( 7 ) Use the above definition (1) to write the Laplace transform of the equation ( 7 ) . s 2 L { x } − s x ( 0 ) − x ′ ( 0 ) + 2 [ s L { x } − x ( 0 ) ] + 2 [ s 2 L { y } − s y ( 0 ) − y ′ ( 0 ) ] = 0 Substitute the value of x ( 0 ) as 0 , y ( 0 ) as 0 , x ′ ( 0 ) as 0 , y ′ ( 0 ) as 0 and L { x } in the above equation

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4, Problem 42CR

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The expression for x(t) and y(t) for the differential equations x″+y″=e2t and 2x′+y″=−e2t by Laplace transform.

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Chapter 4, Problem 41CR

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Chapter 4 Solutions

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The expression for x ( t ) and y ( t ) for the differential equations x ″ + y ″ = e 2 t and 2 x ′ + y ″ = − e 2 t by Laplace transform.

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The expression for x(t) and y(t) for the differential equations x″+y″=e2t and 2x′+y″=−e2t by Laplace transform.

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Chapter 4 Solutions