Explanation Given: The differential system in matrix notation is X ′ = ( 2 5 1 − 5 − 6 4 0 0 2 ) X . Calculation: The differential system X ′ = ( 2 5 1 − 5 − 6 4 0 0 2 ) X is equivalent to X ′ = A X . Here, the matrix are defined as A = ( 2 5 1 − 5 − 6 4 0 0 2 ) , X = ( x y z ) and X ′ = ( d x d t d y d t d z d t ) . The characteristic equation for the above matrix A is det ( A − λ I ) = 0 . det ( A − λ I ) = 0 | 2 − λ 5 1 − 5 − 6 − λ 4 0 0 2 − λ | = 0 ( 2 − λ ) [ ( − 6 − λ ) ( 2 − λ ) − 0 ] − 5 [ − 5 ( 2 − λ ) − 0 ] + 1 ( 0 ) = 0 ( Expand along row 1 ) ( 2 − λ ) ( ( − 6 − λ ) ( 2 − λ ) + 25 ) = 0 ( 2 − λ ) ( λ 2 + 4 λ + 13 ) = 0 Solve the above characteristic equation as follows: 2 − λ = 0 or λ 2 + 4 λ + 13 = 0 Linear equation gives the value of λ = 2 . The quadratic equation λ 2 + 4 λ + 13 = 0 is solved as follows: λ = − ( 4 ) ± ( 4 ) 2 − 4 ( 1 ) ( 13 ) 2 ( 1 ) = − 4 ± 16 − 52 2 = − 4 ± 6 i 2 = − 2 ± 3 i Therefore, the Eigen values of the coefficient matrix are λ 1 = 2 , λ 2 = − 2 + 3 i and λ 3 = − 2 − 3 i . Now, to obtain the Eigenvector corresponding to the Eigen value λ 1 = 2 , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − ( 2 ) I ) K = 0 ( 2 − 2 5 1 − 5 − 6 − 2 4 0 0 2 − 2 ) ( x y z ) = ( 0 0 0 ) ( 0 5 1 − 5 − 8 4 0 0 0 ) ( x y z ) = ( 0 0 0 ) Write the above matrix equation in linear equations form as shown below. 5 y + z = 0 ...... ( 1 ) − 5 x − 8 y + 4 z = 0 ...... ( 2 ) Take y = − 5 and substitute it in equation (1) and obtain the value for z as shown below. 5 ( − 5 ) + z = 0 − 25 + z = 0 z = 25 Substitute y = − 5 and z = 25 in the equation (2) and obtain the value for x as shown below. − 5 x − 8 ( − 5 ) + 4 ( 25 ) = 0 − 5 x + 40 + 100 = 0 − 5 x = − 140 x = 28 Therefore, the first Eigenvector corresponds to the Eigen value λ 1 = 2 is as follows: K 1 = ( 28 − 5 25 ) Therefore, the first corresponding solution of the differential equation is as follows: X 1 = K 1 e λ t X 1 = ( 28 − 5 25 ) e 2 t Similarly, to obtain the Eigenvector corresponding to the Eigen value λ 2 = − 2 + 3 i , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − ( − 2 + 3 i ) I ) K = 0 ( 2 − ( − 2 + 3 i ) 5 1 − 5 − 6 − ( − 2 + 3 i ) 4 0 0 2 − ( − 2 + 3 i ) ) ( x y z ) = ( 0 0 0 ) ( 4 − 3 i 5 1 − 5 − 4 − 3 i 4 0 0 4 − 3 i ) ( x y z ) = ( 0 0 0 ) Simplify the above matrix equation by the use of Gauss Jordan elimination method. ( 4 − 3 i 5 1 0 0 24 5 + i 3 5 0 0 4 − 3 i ) ( x y z ) = ( 0 0 0 ) [ R 2 → R 2 − ( − 4 5 − i 3 5 ) R 1 ] ( 4 − 3 i 5 1 0 0 24 5 + i 3 5 0 0 1 ) ( x y z ) = ( 0 0 0 ) [ R 3 → ( 4 25 + i 3 25 ) R 3 ] ( 4 − 3 i 5 0 0 0 0 0 0 1 ) ( x y z ) = ( 0 0 0 ) [ R 1 → R 1 − R 3 ] Solve the above matrix equation and it gives ( 4 − 3 i ) x + 5 y = 0 and z = 0

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

11th Edition

ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

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Chapter 8.2, Problem 45E

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The general solution of the differential system X′=(251−5−64002)X.

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Chapter 8.2, Problem 44E

Chapter 8.2, Problem 46E

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Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

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The general solution of the differential system X ′ = ( 2 5 1 − 5 − 6 4 0 0 2 ) X .

Solution Summary: The author calculates the differential system Xprime =(cc2& 5& 1 -5& -6& 4

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The general solution of the differential system X′=(251−5−64002)X.

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