Explanation Given: The given non-homogeneous system is, X ′ = ( − 1 5 − 1 1 ) X + ( sin t − 2 cos t ) . Calculation: For the given system, the coefficient matrix is A = ( − 1 5 − 1 1 ) and F ( t ) = ( sin t − 2 cos t ) . Since, the characteristic equation for the matrix A is | ( A − λ I ) | = 0 where, I is the Identity matrix. Substitute the values of matrix A and I in above expression. | − 1 − λ 5 − 1 1 − λ | = 0 − ( 1 + λ ) ( 1 − λ ) + 5 = 0 λ 2 + 4 = 0 Therefore, the Eigen values of the above system is λ 1 = 2 i and λ 2 = − 2 i . Now, to obtain the Eigen vector K 1 corresponding to the Eigen value λ 1 = 2 i . Solve the equation ( A − λ 1 I ) K 1 = 0 where, K 1 = ( x 1 y 1 ) . Substitute the values of A , λ ` 1 and K 1 in the above equation gives, [ ( − 1 5 − 1 1 ) − ( 2 i 0 0 2 i ) ] ⋅ ( x 1 y 1 ) = ( 0 0 ) ( − 1 − 2 i 5 − 1 1 − 2 i ) ⋅ ( x 1 y 1 ) = ( 0 0 ) Applying the row transformation R 2 → R 2 − ( 1 5 − i 2 5 ) R 1 in above system, ( − 1 − 2 i 5 0 0 ) ⋅ ( x 1 y 1 ) = ( 0 0 ) ( − ( 1 + 2 i ) x 1 + 5 y 1 0 ) = ( 0 0 ) Solving above equations gives, y 1 = ( 1 + 2 i ) 5 x 1 Assume that x 1 = n for some scalar n which implies y 1 = ( 1 + 2 i ) 5 n . Therefore, the value of ( x 1 y 1 ) is, ( x 1 y 1 ) = ( n ( 1 + 2 i ) 5 n ) = n ( 1 ( 1 + 2 i ) 5 ) = n ( 5 ( 1 + 2 i ) ) Therefore the eigenvector K 1 corresponding to the Eigen value λ 1 = 2 i is ( 5 ( 1 + 2 i ) ) . Now, to obtain the Eigen vector K 2 corresponding to the Eigen value λ 2 = − 2 i , solve the equation ( A − λ 2 I ) K 2 = 0 where K 2 = ( x 2 y 2 ) . Substitute the values of A , λ ` 2 and K 2 in the above equation gives, [ ( − 1 5 − 1 1 ) − ( − 2 i 0 0 − 2 i ) ] ⋅ ( x 2 y 2 ) = ( 0 0 ) ( − 1 + 2 i 5 − 1 1 + 2 i ) ⋅ ( x 2 y 2 ) = ( 0 0 ) , Applying the row transformation R 2 → R 2 − ( 1 5 + i 2 5 ) R 1 in above system, ( − 1 + 2 i 5 0 0 ) ⋅ ( x 2 y 2 ) = ( 0 0 ) ( ( − 1 + 2 i ) x 2 + 5 y 2 0 ) = ( 0 0 ) Solving above equations gives, y 2 = ( 1 − 2 i ) 5 x 2 Assume that x 2 = m for some scalar m which implies y 2 = ( 1 − 2 i ) 5 m . Therefore, the value of ( x 2 y 2 ) is, ( x 2 y 2 ) = ( m ( 1 − 2 i ) 5 m ) = m ( 1 ( 1 − 2 i ) 5 ) = m ( 5 ( 1 − 2 i ) ) Therefore the eigenvector K 2 corresponding to the Eigen value λ 2 = − 2 i is ( 5 ( 1 − 2 i ) )

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

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 8.3, Problem 6E

To determine

To solve: The given non-homogeneous system using the method of undetermined coefficient.

Expert Solution & Answer

Chapter 8 Solutions

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

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 given non-homogeneous system using the method of undetermined coefficient.

Solution Summary: The author explains how to solve the given non-homogeneous system using the method of undetermined coefficient.

Concept explainers

Videos

To solve: The given non-homogeneous system using the method of undetermined coefficient.

Chapter 8 Solutions