Explanation Given: The differential system in matrix notation is X ′ = ( 4 0 1 0 6 0 − 4 0 4 ) X . Calculation: The differential system X ′ = ( 4 0 1 0 6 0 − 4 0 4 ) X is equivalent to X ′ = A X . Here, the matrix are defined as A = ( 4 0 1 0 6 0 − 4 0 4 ) , 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 | 4 − λ 0 1 0 6 − λ 0 − 4 0 4 − λ | = 0 ( 4 − λ ) [ ( 6 − λ ) ( 4 − λ ) ] + 4 ( 6 − λ ) = 0 ( Expand along row 1 ) ( 6 − λ ) ( λ 2 − 8 λ + 20 ) = 0 Solve the above characteristic equation as follows: 6 − λ = 0 or λ 2 − 8 λ + 20 = 0 Linear equation gives the value of λ = 6 . The quadratic equation λ 2 − 8 λ + 20 = 0 is solved as follows: λ = − ( − 8 ) ± ( − 8 ) 2 − 4 ( 1 ) ( 20 ) 2 ( 1 ) = 8 ± 64 − 80 2 = 8 ± 4 i 2 = 4 ± 2 i Therefore, the Eigen values of the coefficient matrix are λ 1 = 6 , λ 2 = 4 + 2 i and λ 3 = 4 − 2 i . Now, to obtain the Eigenvector corresponding to the Eigen value λ 1 = 6 , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − ( 6 ) I ) K = 0 ( 4 − 6 0 1 0 6 − 6 0 − 4 0 4 − 6 ) ( x y z ) = ( 0 0 0 ) ( − 2 0 1 0 0 0 − 4 0 − 2 ) ( x y z ) = ( 0 0 0 ) Simplify the above matrix equation by the use of Gauss Jordan elimination method. ( − 4 0 − 2 0 0 0 − 2 0 1 ) ( x y z ) = ( 0 0 0 ) [ R 1 ↔ R 3 ] ( − 4 0 − 2 0 0 0 0 0 2 ) ( x y z ) = ( 0 0 0 ) [ R 3 → R 3 − 1 2 R 1 ] ( − 4 0 0 0 0 0 0 0 2 ) ( x y z ) = ( 0 0 0 ) [ R 1 → R 1 + R 3 ] ( − 4 0 0 0 0 0 0 0 2 ) ( x y z ) = ( 0 0 0 ) [ R 1 → R 1 + R 3 ] Further, simplify the above matrix equation. ( 1 0 0 0 0 0 0 0 1 ) ( x y z ) = ( 0 0 0 ) [ R 1 → 1 − 4 R 1 R 3 → 1 2 R 3 ] Solve the above matrix equation and it gives x = 0 and z = 0 and y is independent. Therefore, take y = 1 and the first Eigenvector corresponds to the Eigen value λ 1 = 6 is as follows: K 1 = ( 0 1 0 ) Therefore, the first corresponding solution of the differential equation is as follows: X 1 = K 1 e λ 1 t X 1 = ( 0 1 0 ) e 6 t Similarly, to obtain the Eigenvector corresponding to the Eigen value λ 2 = 4 + 2 i , substitute the value of Eigen value in the equation ( A − λ I ) K = 0 . ( A − ( 4 + 2 i ) I ) K = 0 ( 4 − ( 4 + 2 i ) 0 1 0 6 − ( 4 + 2 i ) 0 − 4 0 4 − ( 4 + 2 i ) ) ( x y z ) = ( 0 0 0 ) ( − 2 i 0 1 0 2 − 2 i 0 − 4 0 − 2 i ) ( x y z ) = ( 0 0 0 ) Simplify the above matrix equation by the use of Gauss Jordan elimination method

11th Edition

ISBN: 9781337652469

Author: ZILL

Publisher: CENGAGE L

See similar textbooks

Videos

Question

Chapter 8.2, Problem 44E

To determine

The general solution of the differential system X′=(401060−404)X.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.2, Problem 43E

Chapter 8.2, Problem 45E

Students have asked these similar questions

Question 6 Find the solution of the exponential equation 100(1.07)² = 500, 000 in terms of logarithms, or correct to four decimal places. t = Question 7 Solve the equation.

I need help on 10

|x6|= 5 The distance between x and is spaces on the number line, in either direction. Next Part

Chapter 8 Solutions

FIRST CRSE.IN DIFF.EQUAT..-ACCESS

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

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

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

Videos

The general solution of the differential system X′=(401060−404)X.

Want to see the full answer?

Chapter 8 Solutions