Explanation Procedure used: The procedure to solve the system using variation of parameter: Find the eigenvalue and eigenvector for matrix A . Let Φ ( t ) be a fundamental matrix for the corresponding homogenous system of linear system X ′ = A X + F ( t ) . Now use the eigenvector to find the fundamental matrix Φ ( t ) . The particular solution has the form, X p = Φ ( t ) U ( t ) , where U ( t ) = ( u 1 ( t ) u 2 ( t ) ⋮ u n ( t ) ) . Find U ( t ) using relation U ( t ) = ∫ Φ − 1 ( t ) F ( t ) d t . Finally calculate particular solution as, X p ( t ) = Φ − 1 ( t ) ∫ Φ − 1 ( t ) F ( t ) d t . The general solution of the equation is X p ( t ) = X ( t ) + X p ( t ) . Calculation: Consider differential equation X ′ = ( 0 1 − 3 − 4 ) X + ( 0 sin t ) The characteristic matrix is A = ( 0 1 − 3 − 4 ) . Write characteristic polynomial equation for matrix A . P ( λ ) = | 0 − λ 1 − 3 − 4 − λ | Expand the determinant. P ( λ ) = − λ ( − 4 − λ ) + 3 = λ 2 + 4 λ + 3 = ( λ + 1 ) ( λ + 3 ) λ = − 1 , and − 3 Find the eigenvector v corresponding to eigenvalue − 1 . | A − λ I | v = 0 ( 0 + 1 1 − 3 − 4 + 1 ) ( v 1 v 2 ) = ( 0 0 ) ( 1 1 − 3 − 3 ) ( v 1 v 2 ) = ( 0 0 ) v 1 + v 2 = 0 The eigenvector v corresponding to eigenvalue λ = − 1 is v = ( 1 − 1 ) . The solution of system of equation is X 1 ( t ) = c 1 e − t ( 1 − 1 ) . Find the eigenvector u corresponding to eigenvalue − 3 . | A − λ I | v = 0 ( 0 + 3 1 − 3 − 4 + 3 ) ( v 1 v 2 ) = ( 0 0 ) ( 3 1 − 3 − 1 ) ( v 1 v 2 ) = ( 0 0 ) 3 v 1 + v 2 = 0 The eigenvector u corresponding to eigenvalue λ = − 3 is v = ( 3 − 1 ) . The solution of system of equation is X 2 ( t ) = c 1 e − 3 t ( 3 − 1 ) . The general solution of equation is X ( t ) = c 1 e − t ( 1 − 1 ) + c 2 e − 3 t ( 3 − 1 ) . Write the fundamental matrix for non-homogenous system X ′ = A X + F ( t ) . Φ ( t ) = ( e − t 3 e − 3 t − e − t − e − 3 t ) The determinant of above matrix Φ ( t ) is, | Φ ( t ) | = | e − t 3 e − 3 t − e − t − e − 3 t | = 2 e − 4 t The inverse of the fundamental matrix is, Φ − 1 ( t ) = 1 2 e − 4 t ( − e − 3 t − e − 3 t e − t e − t ) = ( e t 2 − e t 2 e 3 t 2 − e 3 t 2 ) The homogenous solution of system X ′ = A X + F ( t ) is X h = Φ ( t ) U ( t ) , where U ( t ) = ( c 1 c 2 ) . To find the particular solution, apply the condition U ( t ) = ∫ Φ − 1 ( t ) F ( t ) d t

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 6.5, Problem 28E

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The solution of the system of differential equation X′=(01−3−4)X+(0sint) with boundary condition X(0)=(01).

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Chapter 6.5, Problem 27E

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Chapter 6 Solutions

Introductory Differential Equations

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The solution of the system of differential equation X ′ = ( 0 1 − 3 − 4 ) X + ( 0 sin t ) with boundary condition X ( 0 ) = ( 0 1 ) .

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The solution of the system of differential equation X′=(01−3−4)X+(0sint) with boundary condition X(0)=(01).

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Chapter 6 Solutions