Explanation Given: The given matrix A and b is, A = [ 2 − 4 3 − 9 − 3 − 9 4 4 3 ] , b = [ e 6 t 1 0 ] . Approach: Let, A ( t ) be an n × n matrix function and b ( t ) be a column n -vector function, both of which are continuous on an interval I . If { x 1 , x 2 , .... , x n } is any linearly independent set of solutions to x ' ( t ) = A ( t ) x ( t ) and X ( t ) = [ x 1 ( t ) , x 2 ( t ) , ..... , x n ( t ) ] , then a particular solution to x ' ( t ) = A ( t ) x ( t ) + b ( t ) is, x p ( t ) = X ( t ) u ( t ) , Where u ( t ) satisfies X ( t ) u ' ( t ) = b ( t ) . Calculation: Consider the given matrix A and b , A = [ 2 − 4 3 − 9 − 3 − 9 4 4 3 ] , b = [ e 6 t 1 0 ] . The characteristic equation of A is given as det ( A − λ I ) = 0 , det ( A − λ I ) = 0 | [ 2 − 4 3 − 9 − 3 − 9 4 4 3 ] − [ λ 0 0 0 λ 0 0 0 λ ] | = 0 | 2 − λ − 4 3 − 9 − 3 − λ − 9 4 4 3 − λ | = 0 Solve the above equation for the eigenvalues ( λ ) of A , | 2 − λ − 4 3 − 9 − 3 − λ − 9 4 4 3 − λ | = 0 ( λ + 3 ) ( λ − 6 ) ( λ + 1 ) = 0 λ = − 3 , 6 , − 1 The eigenvector corresponding to λ 1 = − 3 is v 1 = [ − 2 − 1 2 ] , corresponding to λ 2 = 6 is v 2 = [ − 1 1 0 ] and corresponding to λ 3 = − 1 is v 3 = [ − 1 0 1 ] . The linearly independent solution to x ′ = A x are, x 1 ( t ) = e − 3 t [ − 2 − 1 2 ] , x 2 ( t ) = e 6 t [ − 1 1 0 ] and x 3 ( t ) = e − t [ − 1 0 1 ] . Thus the fundamental matrix for x ' = A x is, X ( t ) = [ − 2 e − 3 t − e 6 t − e − t − e − 3 t e 6 t 0 2 e − 3 t 0 e − t ] . The determinant of X ( t ) is, det X ( t ) = | − 2 e − 3 t − e 6 t − e − t − e − 3 t e 6 t 0 2 e − 3 t 0 e − t | = 2 e − 3 t ( e 5 t ) + e − t [ − 2 e 3 t − e 3 t ] = 2 e 2 t − 2 e 2 t − e 2 t = − e 2 t As, the particular solution to x ' = A x + b is, x p = X u Where, X u ' = b . The component of u ' based on Cramer’s rule is, u 1 ′ ( t ) = | e 6 t − e − 6 t − e − t 1 e 6 t 0 0 0 e − t | det X ( t ) = e 11 t + e 5 t − e 2 t = − e 9 t − e 3 t And,

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 9.11, Problem 28AP

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The particular and the general solution to the given matrix A=[2−43−9−3−9443], b=[e6t10].

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Chapter 9.11, Problem 27AP

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The particular and the general solution to the given matrix A = [ 2 − 4 3 − 9 − 3 − 9 4 4 3 ] , b = [ e 6 t 1 0 ] .

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The particular and the general solution to the given matrix A=[2−43−9−3−9443], b=[e6t10].

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