Explanation Given: The given non-homogeneous system is, X ′ = ( 0 − 1 1 0 ) X + ( sec t 0 ) Calculation: The system is given in the form X ′ = A X + F ( t ) (1). Consider, the characteristic equation of a matrix A as | A − λ I | = 0 , where I is identity matrix. Now, the characteristic equation of the coefficient matrix A . is | ( A − λ I ) | = 0 | − λ − 1 1 − λ | = 0 ( − λ ) 2 + 1 = 0 λ = ± i So, the Eigen values are λ 1 = i , λ 2 = − i , . Now evaluating the Eigen vector for λ 1 = 1 + i ,as ( A − ( i ) I 0 ) = ( − i − 1 0 1 − i 0 ) Apply row operation [ R 1 → ( − 1 ) R 1 ] in above expression. ( A − ( i ) I 0 ) = ( i 1 0 1 − i 0 ) Again apply row operation R 2 → R 2 + i R 1 in above expression. ( A − ( i ) I 0 ) = ( i 1 0 0 0 0 ) On solving, it gives − i K 1 + K 2 = 0 Which on further evaluation becomes i K 1 = − K 2 . Thus, the corresponding Eigen vector is K = ( 1 − i ) . Consider the coefficient matrix A having real values of homogeneous system and K be an Eigen vector corresponding the complex Eigen value λ 1 , then the solution is X c = K e λ 1 t Thus, the value of complementary function X c is X c = ( 1 − i ) e i t = ( 1 − i ) ( cos t + i sin t ) = ( cos t sin t ) + i ( sin t − cos t ) On comparing with equation X c = X 1 + i X 2 , it gives the solution vectors of given system as X 1 = ( cos t sin t ) X 2 = ( sin t − cos t ) Now, the entries of X 1 form the first column of solution matrix ϕ ( t ) and the entries of X 2 form the second column of ϕ ( t ) , which means the value of ϕ ( t ) is ϕ ( t ) = ( cos t sin t sin t − cos t ) Now, evaluating determinant of the above matrix ϕ ( t ) gives det [ ϕ ( t ) ] = | cos t sin t sin t − cos t | = − cos 2 t − sin 2 t = − ( cos 2 t + sin 2 t ) = − 1 Now, the value of ϕ − 1 ( t ) is ϕ − 1 ( t ) = 1 | ϕ ( t ) | ⋅ ( Adj

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 8.3, Problem 23E

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To solve: The given non-homogeneous system by using variation of parameters.

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Chapter 8.3, Problem 22E

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The given non-homogeneous system by using variation of parameters.

Solution Summary: The author explains how to solve the given non-homogeneous system by using variation of parameters. The characteristic equation of a matrix A is left|A-lambda Iright|=

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To solve: The given non-homogeneous system by using variation of parameters.

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