Explanation Given: The non-homogeneous system is d y d t = 5 x + 9 y + 2 d y d t = − x + 11 y + 6 Calculation: Express the given non-homogeneous system in the matrix form as given below. X ′ = ( 5 9 − 1 11 ) X + ( 2 6 ) ...... ( 1 ) The characteristic equation of the coefficient matrix can be expressed as follows: ( A − λ I ) = ( 5 − λ 9 − 1 11 − λ ) Express the determinant of the coefficient matrix and equate to zero to obtain the Eigenvalues. | ( A − λ I ) | = | 5 − λ 9 − 1 11 − λ | ( 5 − λ ) ( 11 − λ ) + 9 = 0 λ 2 − 16 λ + 64 = 0 ( λ − 8 ) 2 = 0 Further simplify to obtain the value of λ . ( λ − 8 ) ( λ − 8 ) = 0 λ = 8 Therefore, from the above results, the two Eigenvalues are λ 1 = 8 and λ 2 = 8 . For λ 1 = 8 , the components of Eigenvector is calculated as follows: | 5 − λ 1 9 − 1 11 − λ 1 | = | 5 − 8 9 − 1 11 − 8 | ( λ 1 = 8 ) = | − 3 9 − 1 3 | − 3 K 1 + 9 K 2 = 0 − K 1 + 3 K 2 = 0 where, K 1 , K 2 are the components of the Eigenvector. From the above expression K 2 = 1 3 K 1 . Therefore, the Corresponding Eigenvector is K = ( 3 1 ) . The complementary solution of the system is calculated as follows: ( A − 8 I ) P = K 1 ( − 3 9 − 1 3 ) ( P 1 P 2 ) = ( 3 1 ) − 3 P 1 + 9 P 2 = 3 − P 1 + 3 P 2 = 1 From equation − P 1 + 3 P 2 = 1 , substitute − 1 + 3 P 2 for P 1 in equation − 3 P 1 + 9 P 2 = 3 . − 3 P 1 + 9 P 2 = 3 − 3 ( − 1 + 3 P 2 ) + 9 P 2 = 3 3 − 9 P 2 + 9 P 2 = 3 1 = 1 Consider P 2 as 1 so the value of P 1 is calculated as follows: − 3 P 1 + 9 ( 1 ) = 3 − 3 P 1 = 3 − 9 P 1 = 6 3 = 2 Therefore, the vector P is ( 2 1 )

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

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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