Explanation Given: The system of differential equations is { d x d t = z d y d t = − z d z d t = y . Calculation: The given system of differential equation in matrix notation is as follows: X ′ = ( 0 0 1 0 0 − 1 0 1 0 ) X Here, the matrix are defined as A = ( 0 0 1 0 0 − 1 0 1 0 ) , 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 | 0 − λ 0 1 0 0 − λ − 1 0 1 0 − λ | = 0 ( − λ ) ( λ 2 + 1 ) = 0 λ ( λ 2 + 1 ) = 0 Solve the above characteristic equation as follows: λ = 0 or λ 2 + 1 = 0 λ = 0 or λ 2 = − 1 λ = 0 or λ = ± i Therefore, the Eigen values of the coefficient matrix are λ 1 = 0 , λ 2 = i and λ 3 = λ 2 ¯ . Now, to obtain the Eigenvector corresponding to the Eigen value λ 1 = 0 , substitute the value of Eigen value in the equation given below. ( A − 0 I ) K = 0 ( 0 − 0 0 1 0 0 − 0 − 1 0 1 0 − 0 ) ( x y z ) = ( 0 0 0 ) ( 0 0 1 0 0 − 1 0 1 0 ) ( x y z ) = ( 0 0 0 ) Write the augmented matrix of the above equation. ( A − ( 0 ) I 0 ) = ( 0 0 1 0 0 0 − 1 0 0 1 0 0 ) Here, the augmented matrix is equivalent to linear equations as shown below. y = 0 z = 0 The value for x is independent, therefore take 1 for x . The first Eigenvector corresponds to the Eigen value λ 1 = 0 is as follows: K 1 = ( 1 0 0 ) Therefore, the first corresponding solution of the differential equation is as follows: X 1 = K 1 e λ 1 t X 1 = ( 1 0 0 ) e 0 t = ( 1 0 0 ) Similarly, to obtain the Eigenvector corresponding to the Eigen value λ 2 = i , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − i I ) K = 0 ( 0 − i 0 1 0 0 − i − 1 0 1 0 − i ) ( x y z ) = ( 0 0 0 ) ( − i 0 1 0 − i − 1 0 1 − i ) ( x y z ) = ( 0 0 0 ) Write the augmented matrix of the above equation and solve. ( A − ( i ) I 0 ) = ( − i 0 1 0 0 − i − 1 0 0 1 − i 0 ) Here, the augmented matrix is equivalent to linear equations as shown below. − i x + z = 0 ...... ( a ) − i y − z = 0 ...... ( b ) y − i z = 0 ...... ( c ) Equation (b) and (c) are equivalent equations. Therefore, it is observed that z is independent variable. Take z = 1 and solve for x and y . Substitute z = 1 and obtain the value for x from the equation (a). − i x + ( 1 ) = 0 − i x = − 1 x = − i ( Multiply by i ) Substitute z = 1 and obtain the value for y from the equation (b). − i y − ( 1 ) = 0 − i y = 1 y = i ( Multiply by i ) The second Eigenvector corresponds to the Eigen value λ 2 = i is as follows: K 2 = ( − i i 1 ) Therefore, the second corresponding solution of the differential equation is as follows: X 2 = K 2 e λ 2 t = ( − i i 1 ) e i t = ( − i i 1 ) ( cos t + i sin t ) = ( − i ( cos t + i sin t ) i ( cos t + i sin t ) ( cos t + i sin t ) ) Further, simplify the above solution

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 10.2, Problem 41E

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The general solution of the system of differential equations {dxdt=zdydt=−zdzdt=y.

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Chapter 10.2, Problem 40E

Chapter 10.2, Problem 42E

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The general solution of the system of differential equations { d x d t = z d y d t = − z d z d t = y .

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