Explanation Given: The non-homogeneous system is X ′ = ( 2 − 1 4 2 ) X + ( sin 2 t 2 cos 2 t ) e 2 t . Calculation: The system is in the form X ′ = A X + F ( t ) . Consider, the characteristic equation of a matrix A as | A − λ I | = 0 . Now, the characteristic equation of the coefficient matrix A is given below. | ( A − λ I ) | = 0 | 2 − λ − 1 4 2 − λ | = 0 ( 2 − λ ) ( 2 − λ ) + 4 = 0 4 − 4 λ + λ 2 + 4 = 0 Further simplify the above equation. λ 2 − 4 λ + 8 = 0 λ = − ( − 4 ) ± ( − 4 ) 2 − ( 4 ) ( 1 ) ( 8 ) ( 2 ) ( 1 ) λ = 4 ± 4 i 2 λ 1 = 2 + 2 i or λ 2 = 2 − 2 i Therefore, the Eigen values are λ 1 = 2 + 2 i , λ 2 = 2 − 2 i . Now, the Eigen vector for λ 1 = 2 + 2 i is given below. ( A − ( 2 + 2 i ) I 0 ) = ( − 2 i − 1 0 4 − 2 i 0 ) Apply row operation R 1 = ( − 1 2 i ) R 1 in above expression. ( A − ( 2 + 2 i ) I 0 ) = ( 1 1 2 i 0 4 − 2 i 0 ) Apply row operation R 2 → R 2 − R 1 in above expression. ( A − ( 2 + 2 i ) I 0 ) = ( 1 − i 2 0 0 0 0 ) Therefore, after solving, the expression K 1 − i 2 K 2 = 0 . Therefore, the value of expression is K 1 = i 2 K 2 . So the corresponding Eigen vector is K = ( − i 2 1 ) . Consider the coefficient matrix A with real values of homogeneous system and K is an Eigen vector corresponding the complex Eigen value λ 1 = 2 + 2 i then the solution is X c = K e λ 1 t . X c = X 1 + X 2 ...... ( 1 ) Substitute ( i 2 1 ) for K and 2 + 2 i for λ 1 in the equation X c = K e λ 1 t as follows: X c = ( i 2 1 ) e ( 2 + 2 i ) t = e 2 t ( i 2 1 ) ( cos 2 t + i sin 2 t ) = ( − sin 2 t 2 cos 2 t ) e 2 t + i ( cos 2 t 2 sin 2 t ) e 2 t On comparing with equation (1) it gives the solution vectors of given system as X 1 = ( − sin 2 t 2 cos 2 t ) e 2 t , X 2 = i ( cos 2 t 2 sin 2 t ) e 2 t The entries in X 1 form the first column of solution matrix ϕ ( t ) , and the entries in X 2 form the second column of ϕ ( t ) . ϕ ( t ) = ( − e 2 t sin 2 t 2 e 2 t cos 2 t 2 e 2 t cos 2 t e 2 t sin 2 t ) Evaluating determinant of the above matrix ϕ ( t ) , det [ ϕ ( t ) ] = | − e 2 t sin 2 t 2 e 2 t cos 2 t 2 e 2 t cos 2 t e 2 t sin 2 t | = − e 4 t sin 2 2 t 2 − e 4 t cos 2 2 t 2 = − e 4 t 2 Now, the value of the value of ϕ − 1 ( t ) is ϕ − 1 ( t ) = 1 | ϕ ( t ) | ⋅ ( Adj . ϕ ( t ) ) ϕ − 1 ( t ) = 1 | ϕ ( t ) | ⋅ [ Adj

Bundle: A First Course in Differential Equations with Modeling Applications, Loose-leaf Version, 11th + WebAssign Printed Access Card for Zill's A ... Applications, 11th Edition, Single-Term

11th Edition

ISBN: 9781337604994

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 8.3, Problem 16E

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The non-homogeneous system X′=(2−142)X+(sin2t2cos2t)e2t using variation of parameter.

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The non-homogeneous system X ′ = ( 2 − 1 4 2 ) X + ( sin 2 t 2 cos 2 t ) e 2 t using variation of parameter.

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The non-homogeneous system X′=(2−142)X+(sin2t2cos2t)e2t using variation of parameter.

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