Explanation Given: The given system is: X ′ ( t ) = [ 3 − 2 1 2 − 1 1 − 4 4 1 ] X ( t ) Procedure used: Suppose n × n constant matrix A has n linearly independent eigenvectors u 1 , u 2 , ⋯ , u n . The eigenvalues can be finding by using characteristics equation | A − r I | = 0 . Here, r denotes the eigenvalues. The corresponding the vector of the eigenvalues are called eigenvectors. It can be found by substitute the eigenvalues in the formula ( A − r I ) u = 0 . If the eigenvalues are repeated then the solution is: X 1 ( t ) = e r t u 1 , X 2 ( t ) = t e r t u 1 + e r t u 2 and X 3 ( t ) = t 2 2 e r t u 1 + t e r t u 2 + e r t u 3 The general solution is; X ( t ) = X 1 ( t ) + X 2 ( t ) + X 3 ( t ) Calculation: Consider the given matrix: A = [ 3 − 2 1 2 − 1 1 − 4 4 1 ] Substitute the value of matrix A in the characteristic equation and solve for r : | [ 3 − 2 1 2 − 1 1 − 4 4 1 ] − r [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ 3 − 2 1 2 − 1 1 − 4 4 1 ] − [ r 0 0 0 r 0 0 0 r ] | = 0 | [ 3 − r − 2 1 2 − 1 − r 1 − 4 4 1 − r ] | = 0 ( 3 − r ) ⋅ | − 1 − r 1 4 1 − r | + 2 ⋅ | 2 1 − 4 1 − r | + 1. | 2 − 1 − r − 4 4 | = 0 Further simplify as: ( 3 − r ) ⋅ [ ( − 1 − r ) ( 1 − r ) − 4 ] + 2 ⋅ [ 2 ( 1 − r ) + 4 ] + 1. [ 8 − 4 ( 1 + r ) ] = 0 ( 3 − r ) ⋅ [ − ( 1 − r 2 ) − 4 ] + 2 ⋅ [ 2 − 2 r + 4 ] + [ 8 − 4 − 4 r ] = 0 ( 3 − r ) ⋅ [ r 2 − 5 ] + 12 − 4 r + 4 − 4 r = 0 3 r 2 − 15 − r 3 + 5 r + 16 − 8 r = 0 Further simplify as: − r 3 + 3 r 2 − 3 r + 1 = 0 − ( r − 1 ) 3 = 0 The factor is: ( r − 1 ) 3 = 0 r = 1 The eigenvalues are r 1 , 2 , 3 = 1 . Therefore, the eigenvalues are repeated. Find the eigenvector corresponding to eigenvalue r 1 = 1 by using the formula ( A − r I ) u = 0 :

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 9.5, Problem 38E

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Chapter 9.5, Problem 37E

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Solution Summary: The author explains the general solution for the given system. The eigenvalues can be found by using characteristics equation left|A-r

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Chapter 9 Solutions

To find:

The general solution for the given system.