Explanation Given: The linear system is X ′ = ( 0 2 1 1 1 − 2 2 2 − 1 ) X . Calculation: The system is in the form X ′ = A X . Write the characteristic equation of the coefficient matrix A . ( A − λ I ) = ( − λ 2 1 1 1 − λ − 2 2 2 − 1 − λ ) For the Eigen values, determinant of the coefficient matrix equate to zero. ( A − λ I ) = | − λ 2 1 1 1 − λ − 2 2 2 − 1 − λ | = ( − λ ) [ ( 1 − λ ) ( − 1 − λ ) − 2 ⋅ ( − 2 ) ] − ( 2 ) [ 1 ⋅ ( − 1 − λ ) − 2 ⋅ ( − 2 ) ] + 1 [ 1 ⋅ 2 − ( 1 − λ ) ⋅ 2 ] = − λ 3 + 3 − λ + 3 + 2 λ = − λ 3 + λ − 6 = 0 Simplify to obtain Eigen values as follows: − ( λ + 2 ) ( λ 2 − 2 λ + 3 ) = 0 ( λ + 2 ) ( λ 2 − 2 λ + 3 ) = 0 λ 1 = − 2 or λ 2 = 1 + 2 i or λ 3 = 1 − 2 i where, λ 1 , λ 2 and λ 3 are the three Eigen values. For λ 1 = − 2 , the Eigen vector is given below. ( A − ( − 2 ) I | 0 ) = ( 2 2 1 0 1 3 − 2 0 2 2 1 0 ) Therefore, the values of Eigen vector are as follows: 2 K 1 + 2 K 2 + K 3 = 0 ...... ( 1 ) K 1 + 3 K 2 + ( − 2 ) K 3 = 0 ...... ( 2 ) Consider the value of K 3 as 4 then the above expression is given below. 2 K 1 + 2 K 2 + 4 = 0 K 1 + 3 K 2 − 8 = 0 Multiply the equation K 1 + 3 K 2 − 8 = 0 by − 2 . 2 K 1 + 2 K 2 + 4 = 0 − 2 K 1 − 6 K 2 + 16 = 0 _ − 4 K 2 + 20 = 0 The expression − 4 K 2 + 20 = 0 is solved as follows: − 4 K 2 + 20 = 0 K 2 = 20 4 K 2 = 5 Therefore, the value of K 2 is 5 . Substitute 5 for K 2 in equation 2 K 1 + 2 K 2 + 4 = 0 to obtain the value of K 1 . 2 K 1 + 2 ( 5 ) + 4 = 0 2 K 1 + 10 + 4 = 0 K 1 = − 14 2 K 1 = − 7 Therefore, the value of K 1 is − 7 . The Corresponding Eigen vector is K 1 = ( − 7 5 4 ) . Therefore, the Corresponding solution vector is X 1 = ( − 7 5 4 ) e − 2 t For λ 2 = 1 + 2 i , the Eigen vector is given below. ( A + ( 2 i ) I | 0 ) = ( − 1 − 2 i 2 1 0 1 − 2 i − 2 0 2 2 − 2 − 2 i 0 ) The equations are as follows: ( − 1 − 2 i ) K 1 + 2 K 2 + K 3 = 0 K 1 − 2 i K 2 − 2 K 3 = 0 2 K 1 + 2 K 2 + ( − 2 − 2 i ) K 3 = 0 Subtract the equation ( − 1 − 2 i ) K 1 + 2 K 2 + K 3 = 0 and 2 K 1 + 2 K 2 + ( − 2 − 2 i ) K 3 = 0 as follows: ( − 1 − 2 i ) K 1 + 2 K 2 + K 3 = 0 2 K 1 + 2 K 2 + ( − 2 − 2 i ) K 3 = 0 − − + _ ( − 1 − 2 i − 2 ) K 1 + ( 1 + 2 + 2 i ) K 3 = 0

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, Problem 10RE

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The solution of the linear system X′=(02111−222−1)X.

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

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

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The solution of the linear system X′=(02111−222−1)X.

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