Explanation Given: The system of differential equations is { d x d t = 2 x + y + 2 z d y d t = 3 x + 6 z d z d t = − 4 x − 3 z . Calculation: The given system of differential equation in matrix notation is as follows: X ′ = ( 2 1 2 3 0 6 − 4 0 − 3 ) X Here, the matrix are defined as A = ( 2 1 2 3 0 6 − 4 0 − 3 ) , 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 1 2 3 0 − λ 6 − 4 0 − 3 − λ | = 0 − 4 ( 6 + 2 λ ) + ( − 3 − λ ) ( ( − λ ) ( 2 − λ ) − 3 ) = 0 ( Expand along row 3 ) − 4 ( 6 + 2 λ ) − ( 3 + λ ) ( − 2 λ + λ 2 − 3 ) = 0 Further, simplify the above equation. − 4 ( 2 ) ( 3 + λ ) − ( 3 + λ ) ( − 2 λ + λ 2 − 3 ) = 0 ( 3 + λ ) { − 8 + 2 λ − λ 2 + 3 } = 0 − ( 3 + λ ) { 8 − 2 λ + λ 2 − 3 } = 0 ( 3 + λ ) ( λ 2 − 2 λ + 5 ) = 0 Solve the above characteristic equation as follows: 3 + λ = 0 or λ 2 − 2 λ + 5 = 0 Linear equation gives the value of λ = − 3 . The quadratic equation λ 2 − 2 λ + 5 = 0 is solved as follows: λ = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( 5 ) 2 ( 1 ) = 2 ± 4 − 20 2 = 2 ± 4 i 2 = 1 ± 2 i Therefore, the Eigen values of the coefficient matrix are λ 1 = − 3 , λ 2 = 1 + 2 i and λ 3 = 1 − 2 i . Now, to obtain the Eigenvector corresponding to the Eigen value λ 1 = − 3 , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − ( − 3 ) I ) K = 0 ( 5 1 2 3 3 6 − 4 0 0 ) ( x y z ) = ( 0 0 0 ) Here, the matrix equation is equivalent to linear equations as shown below. 5 x + y + 2 z = 0 ...... ( 1 ) 3 x + 3 y + 6 z = 0 ...... ( 2 ) − 4 x = 0 ...... ( 3 ) From equation (3), the value for x is 0 . Substitute 0 for x in equation (1) and (2). y + 2 z = 0 ...... ( 4 ) 3 y + 6 z = 0 ...... ( 5 ) It is observed that both the above equations are equivalent. Therefore, take z = 1 in the equation (4) and obtain the value of y = − 2 . So, the first Eigenvector corresponds to the Eigen value λ 1 = − 3 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 − 3 t Similarly, to obtain the Eigenvector corresponding to the Eigen value λ 2 = 1 + 2 i , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − ( 1 + 2 i ) I ) K = 0 ( 2 − ( 1 + 2 i ) 1 2 3 − ( 1 + 2 i ) 6 − 4 0 − 3 − ( 1 + 2 i ) ) ( x y z ) = ( 0 0 0 ) ( 1 − 2 i 1 2 3 − ( 1 + 2 i ) 6 − 4 0 − 4 − 2 i ) ( x y z ) = ( 0 0 0 ) Here, the matrix equation is equivalent to linear equations as shown below. ( 1 − 2 i ) x + y + 2 z = 0 ...... ( 6 ) 3 x − ( 1 + 2 i ) y + 6 z = 0

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Chapter 8.2, Problem 42E

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The general solution of the system of differential equations {dxdt=2x+y+2zdydt=3x+6zdzdt=−4x−3z.

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Chapter 8.2, Problem 41E

Chapter 8.2, Problem 43E

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The general solution of the system of differential equations { d x d t = 2 x + y + 2 z d y d t = 3 x + 6 z d z d t = − 4 x − 3 z .

Solution Summary: The author explains the general solution of the system of differential equations, which is ldx.

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The general solution of the system of differential equations {dxdt=2x+y+2zdydt=3x+6zdzdt=−4x−3z.

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