Explanation Given: The given system is X ′ = ( 8 − 9 1 − 2 ) X − 8 ( 1 − 1 ) e 7 t . Procedure used: First find the eigenvalues using | A − λ I | = 0 . A solution of λ to this equation gives eigenvalues and vector v satisfying ( A − λ I ) v = 0 is an eigenvectors corresponds to λ . From the eigenvalues and eigenvectors, X c ( t ) = c 1 v 1 e λ 1 t + c 2 v 2 e λ 2 t + ⋯ + c n v n e λ n t . The particular solution for the system is X p ( t ) = Φ ( t ) ∫ Φ − 1 ( t ) F ( t ) d t . Here, Φ ( t ) is the fundamental matrix. The general solution is, X ( t ) = X c + X p . Calculation: Consider the system X ′ = ( 8 − 9 1 − 2 ) X − 8 ( 1 − 1 ) e 7 t . Apply the formula to find the eigenvalues: | A − λ I | = | ( 8 − 9 1 − 2 ) − λ ( 1 0 0 1 ) | = | ( 8 − 9 1 − 2 ) − ( λ 0 0 λ ) | = | ( 8 − λ − 9 1 − 2 − λ ) | = ( 8 − λ ) ( − 2 − λ ) + 9 Further simplify as: ( 8 − λ ) ( − 2 − λ ) + 9 = − 16 − 8 λ + 2 λ + λ 2 + 9 = λ 2 − 6 λ − 7 = λ 2 − 7 λ + λ − 7 = ( λ − 7 ) ( λ + 1 ) Equate to zero and find λ : ( λ − 7 ) ( λ + 1 ) = 0 ( λ − 7 ) = 0 λ = 7 ( λ + 1 ) = 0 λ = − 1 The eigenvalues are λ 1 = − 1 and λ 2 = 7 . The eigenvector corresponds to λ 1 = − 1 : | A − λ I | = ( 8 − 9 1 − 2 ) − ( − 1 ) ( 1 0 0 1 ) = ( 8 − 9 1 − 2 ) + ( 1 0 0 1 ) = ( 9 − 9 1 − 1 ) Further reduce as: ( 9 − 9 1 − 1 ) → R 2 → R 2 − 1 9 R 1 ( 9 − 9 0 0 ) It indicate that, 9 x 1 − 9 y 1 = 0 . Choose x 1 = 1 , then y 1 = 1 gives v 1 = ( 1 1 ) . The solution X 1 is X 1 = ( 1 1 ) e − t . The eigenvector corresponds to λ 2 = 7 : | A − λ I | = ( 8 − 9 1 − 2 ) − ( 7 ) ( 1 0 0 1 ) = ( 8 − 9 1 − 2 ) − ( 7 0 0 7 ) = ( 1 − 9 1 − 9 ) Further reduce as: ( 1 − 9 1 − 9 ) → R 2 → R 2 − R 1 ( 1 − 9 0 0 ) It indicate that, x 2 − 9 y 2 = 0 . Choose x 2 = 9 , then y 2 = 1 gives v 2 = ( 9 1 ) . The solution X 2 is, X 2 = ( 9 1 ) e 7 t

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 6 Solutions

Introductory Differential Equations

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