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 ) )

11th Edition

ISBN: 9781337652469

Author: ZILL

Publisher: CENGAGE L

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

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

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