Explanation Given: The differential system in matrix notation is X ′ = ( 1 − 1 2 − 1 1 0 − 1 0 1 ) X . Calculation: The differential system X ′ = ( 1 − 1 2 − 1 1 0 − 1 0 1 ) X is equivalent to X ′ = A X . Here, the matrix are defined as A = ( 1 − 1 2 − 1 1 0 − 1 0 1 ) , 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 | 1 − λ − 1 2 − 1 1 − λ 0 − 1 0 1 − λ | = 0 ( 1 − λ ) ( 1 − λ ) ( 1 − λ ) + ( − 1 ) ( 1 − λ ) − 2 ( − 1 ) ( 1 − λ ) = 0 ( Expand along row 1 ) ( 1 − λ ) 3 − ( 1 − λ ) + 2 ( 1 − λ ) = 0 Further, simplify the above equation. ( 1 − λ ) 3 + ( 1 − λ ) = 0 ( 1 − λ ) { λ 2 − 2 λ + 2 } = 0 Solve the above characteristic equation as follows: 1 − λ = 0 or λ 2 − 2 λ + 2 = 0 Linear equation gives the value of λ = 1 . The quadratic equation λ 2 − 2 λ + 2 = 0 is solved as follows: λ = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( 2 ) 2 ( 1 ) = 2 ± 4 − 8 2 = 2 ± 2 i 2 = 1 ± i Therefore, the Eigen values of the coefficient matrix are λ 1 = 1 , λ 2 = 1 + i and λ 3 = 1 − i . Now, to obtain the Eigenvector corresponding to the Eigen value λ 1 = 1 , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − ( 1 ) I ) K = 0 ( 1 − 1 − 1 2 − 1 1 − 1 0 − 1 0 1 − 1 ) ( x y z ) = ( 0 0 0 ) ( 0 − 1 2 − 1 0 0 − 1 0 0 ) ( x y z ) = ( 0 0 0 ) Here, the matrix equation is equivalent to linear equations as shown below. − y + 2 z = 0 ...... ( a ) − x = 0 ...... ( b ) − x = 0 ...... ( c ) From equation (c) or (b), the value for x is 0 . Therefore, take z = 1 in the equation (a) and obtain the value of y = 2 . So, the first Eigenvector corresponds to the Eigen value λ 1 = 1 is as follows: K 1 = ( 0 2 1 ) Therefore, the first corresponding solution of the differential equation is as follows: X 1 = K 1 e λ 1 t X 1 = ( 0 2 1 ) e t Similarly, to obtain the Eigenvector corresponding to the Eigen value λ 2 = 1 + i , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − ( 1 + i ) I ) K = 0 ( − i − 1 2 − 1 − i 0 − 1 0 − i ) ( x y z ) = ( 0 0 0 ) ( − 2 i 0 2 − 1 − i 0 − 1 0 − i ) ( x y z ) = ( 0 0 0 ) ( R 1 → R 1 + i R 2 ) Here, the matrix equation is equivalent to linear equations as shown below

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

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ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 8.2, Problem 43E

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The general solution of the differential system X′=(1−12−110−101)X.

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Chapter 8.2, Problem 42E

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A First Course in Differential Equations with Modeling Applications (MindTap Course List)

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

Solution Summary: The author explains how the differential system Xprime is equivalent to the matrix A=left.

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The general solution of the differential system X′=(1−12−110−101)X.

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