Explanation Consider the given systems of differential equation as { y ′ 1 = 10 y 1 − 10 y 2 − 4 y 3 y ′ 2 = − 10 y 1 + y 2 − 14 y 3 y ′ 3 = − 4 y 1 − 14 y 2 − 2 y 3 . The coefficient matrix of the corresponding systems of differential equations is A = [ 10 − 10 − 4 − 10 1 − 14 − 4 − 14 − 2 ] . Note that, a single vector equation is y ′ = A y , where A is coefficient matrix . Hence, a single vector equation is y ′ = [ 10 − 10 − 4 − 10 1 − 14 − 4 − 14 − 2 ] y , where y = [ y 1 y 2 y 3 ] . Characteristic equation of the coefficient matrix A = [ 10 − 10 − 4 − 10 1 − 14 − 4 − 14 − 2 ] is det ( A − λ I ) = 0 , where λ eigenvalue and I is 3 × 3 identity matrix . Simplify for the det ( A − λ I ) = 0 using the coefficient matrix as follows. det ( A − λ I ) = 0 det ( [ 10 − 10 − 4 − 10 1 − 14 − 4 − 14 − 2 ] − λ [ 1 0 0 0 1 0 0 0 1 ] ) = 0 det ( [ 10 − λ − 10 − 4 − 10 1 − λ − 14 − 4 − 14 − 2 − λ ] ) = 0 ( 10 − λ ) { ( 1 − λ ) ( − 2 − λ ) − 196 } + 10 { − 10 ( − 2 − λ ) − 56 } − 4 { 140 + 4 ( 1 − λ ) } = 0 Further simplified as, ( 10 − λ ) { λ 2 + λ − 198 } + 10 { 20 + 10 λ − 56 } − 4 { 140 + 4 − 4 λ } = 0 ( 10 − λ ) { λ 2 + λ − 198 } + 100 λ − 360 − 576 + 16 λ = 0 λ 2 ( λ − 9 ) − 324 ( λ − 9 ) = 0 ( λ − 9 ) ( 324 − λ 2 ) = 0 λ − 9 = 0 or 324 − λ 2 = 0 λ = 9 or λ = ± 18 Compute the eigenvector x ( 1 ) = [ x 1 x 2 x 3 ] corresponding to the eigenvalue λ = 9 as follows. A x − ( − 9 ) x = 0 [ 10 − λ − 10 − 4 − 10 1 − λ − 14 − 4 − 14 − 2 − λ ] [ x 1 x 2 x 3 ] = [ 0 0 0 ] [ ( 10 − λ ) x 1 − 10 x 2 − 4 x 3 − 10 x 1 + ( 1 − λ ) x 2 − 14 x 3 − 4 x 1 − 14 x 2 − ( 2 + λ ) x 3 ] = [ 0 0 0 ] ( 10 − λ ) x 1 − 10 x 2 − 4 x 3 = 0 , − 10 x 1 + ( 1 − λ ) x 2 − 14 x 3 = 0 and − 4 x 1 − 14 x 2 − ( 2 + λ ) x 3 = 0 Consider the general eigenvector equation for eigenvalue λ as ( 10 − λ ) x 1 − 10 x 2 − 4 x 3 = 0

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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Chapter 4.3, Problem 9P

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The real general solution of the given systems of differential equation.

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