Explanation Given: The differential equation is d x d t = 4 x − 2 y + 2 U ( t − 1 ) d y d t = 3 x − y + U ( t − 1 ) . And the initial conditions are x ( 0 ) = 0 , y ( 0 ) = 1 2 . Concept: Let f be a function defined for t ≥ 0 then the integral L { f ( t ) } = ∫ 0 ∞ e − s t f ( t ) d t is said to be the Laplace transform of f provided that the integral converges. Calculation: The system of differential equations are d x d t = 4 x − 2 y + 2 U ( t − 1 ) and d y d t = 3 x − y + U ( t − 1 ) . Rewrite the first equation as given below. x ′ = 4 x − 2 y + 2 U ( t − 1 ) ...... ( 1 ) Rewrite the second equation as given below. y ′ = 3 x − y + U ( t − 1 ) ...... ( 2 ) Apply Laplace transform to the above equation (1). x ′ = 4 x − 2 y + 2 U ( t − 1 ) L { x ′ } = 4 L { x } − 2 L { y } + 2 L { U ( t − 1 ) } ...... ( 3 ) Apply Laplace transform to the above equation (2). y ′ = 3 x − y + U ( t − 1 ) L { y ′ } = 3 L { x } − L { y } + L { U ( t − 1 ) } ...... ( 4 ) According to the Laplace transform of x ′ can be written as follows: L { x ′ } = s X ( s ) − x ( 0 ) ...... ( 5 ) Now substitute 0 for x ( 0 ) in equation (5). L { x ′ } = s X ( s ) − 0 = s X ( s ) And assume the values as L { x } = X ( s ) , L { y } = Y ( s ) . Now, consider the function is L { U ( t − 1 ) } . Compare the function L { U ( t − 1 ) } with the general form L { U ( t − a ) } = e − a s s . a = 1 Therefore, after comparing, the value of a is 1 then L { U ( t − 1 ) } = e − s s . Substitute e − s s for L { U ( t − 1 ) } in equation (3). L { x ′ } = 4 L { x } − 2 L { y } + 2 L { U ( t − 1 ) } s X ( s ) = 4 X ( s ) − 2 Y ( s ) + 2 e − s s Further simplify the above equation. ( s − 4 ) X ( s ) + 2 Y ( s ) = 2 e − s s ...... ( 6 ) Similarly, the equation (4) becomes as follows: ( s + 1 ) Y ( s ) − 3 X ( s ) = 2 e − s + S 2 s ...... ( 7 ) Multiply equation (7) by ( s − 4 ) and equation (6) by 3 and then add these two equations. ( s − 4 ) ( s + 1 ) Y ( s ) − 3 ( s − 4 ) X ( s ) + 3 ( s − 4 ) X ( s ) + 6 Y ( s ) = 6 e − s s + ( s − 4 ) 2 e − s + S 2 s Further simplify the above equation. Y ( s ) = s − 4 + 2 e − s 2 s ( s − 3 ) Apply inverse Laplace transform both sides of the above equation. L − 1 { Y ( s ) } = L − 1 { s − 4 + 2 e − s 2 s ( s − 3 ) } y ( t ) = L − 1 { s − 4 2 s ( s − 3 ) } + L − 1 { 2 e − s 2 s ( s − 3 ) } = 1 2 L − 1 { s − 4 s ( s − 3 ) } + L − 1 { 1 s ( s − 3 ) e − s } Now, the partial fraction of s − 4 s ( s − 3 ) is calculated as follows: s − 4 s ( s − 3 ) = A s + B s − 3 s − 4 s ( s − 3 ) = A ( s − 3 ) + B s s ( s − 3 ) s − 4 = A ( s − 3 ) + B s

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 7.6, Problem 12E

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The solution of the Laplace transform of the differential equations dxdt=4x−2y+2U(t−1) and dydt=3x−y+U(t−1).

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Chapter 7.6, Problem 11E

Chapter 7.6, Problem 13E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - Prob. 10E

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The solution of the Laplace transform of the differential equations d x d t = 4 x − 2 y + 2 U ( t − 1 ) and d y d t = 3 x − y + U ( t − 1 ) .

Solution Summary: The author explains the Laplace transform of the differential equations dx, and the integral Lleftf (t).

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The solution of the Laplace transform of the differential equations dxdt=4x−2y+2U(t−1) and dydt=3x−y+U(t−1).

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Chapter 7 Solutions