Explanation Given: The given system of differential equations, x ′ 1 = − 12 x 1 − 7 x 2 and x ′ 2 = 16 x 1 + 10 x 2 . Approach: Let, the differential equations are, x ′ 1 = a x 1 + b x 2 x ′ 2 = c x 1 + d x 2 . The given system can be written as, x ′ = A x Here, matrix A is, A = [ a b c d ] The transformed system is, y ′ = ( S − 1 A S ) y Where, x = S y To evaluate the matrix S , it is necessary to find out the eigenvalues and eigenvectors of A . Calculation: Consider the given system of differential equations, x ′ 1 = − 12 x 1 − 7 x 2 x ′ 2 = 16 x 1 + 10 x 2 . The given system can be written as, x ′ = A x [ x ′ 1 x ′ 2 ] = [ − 12 − 7 16 10 ] [ x 1 x 2 ] Here, matrix A is, A = [ − 12 − 7 16 10 ] Consider the matrix A , A = [ − 12 − 7 16 10 ] The characteristic equation of A is given as det ( A − λ I ) = 0 . det ( A − λ I ) = 0 | [ − 12 − 7 16 10 ] − [ λ 0 0 λ ] | = 0 | [ − 12 − λ − 7 16 10 − λ ] | = 0 Solve the above equation for the eigenvalues ( λ ) of A . | − 12 − λ − 7 16 10 − λ | = 0 ( − 12 − λ ) ( 10 − λ ) − ( 16 ) ( − 7 ) = 0 − 120 + 12 λ − 10 λ + λ 2 + 112 = 0 λ 2 + 2 λ − 8 = 0 Further solve the above equation. λ 2 + 4 λ − 2 λ − 8 = 0 λ ( λ + 4 ) − 2 ( λ + 4 ) = 0 ( λ − 2 ) ( λ + 4 ) = 0 λ = 2 , − 4 Therefore, the eigenvalue of A are 2 and − 4 . The eigenvector corresponding to the eigenvalue λ 1 = 2 is determined based on the linear system given as, ( A − λ I ) v = 0 ( [ − 12 − 7 16 10 ] − [ λ 0 0 λ ] ) [ v 1 v 2 ] = [ 0 0 ] [ − 12 − λ − 7 16 10 − λ ] [ v 1 v 2 ] = [ 0 0 ] Substitute 2 for λ in the above equation. [ − 12 − ( 2 ) − 7 16 10 − ( 2 ) ] [ v 1 v 2 ] = [ 0 0 ] [ − 12 − 2 − 7 16 10 − 2 ] [ v 1 v 2 ] = [ 0 0 ] [ − 14 − 7 16 8 ] [ v 1 v 2 ] = [ 0 0 ] The augmented matrix of the above expression is, [ − 14 − 7 16 8 | 0 0 ] Apply R 1 → − 1 7 R 1 .and R 2 → 1 8 R 2 . [ 2 1 2 1 | 0 0 ] Apply R 2 → R 2 − R 1 . [ 2 1 0 0 | 0 0 ] Thus, [ 2 1 0 0 ] [ v 1 v 2 ] = [ 0 0 ] [ 2 v 1 + v 2 0 ] = [ 0 0 ] The equation obtained is, 2 v 1 + v 2 = 0 Solve the above equation for v 2 . v 2 = − 2 v 1 Let, v 1 = k , then v 2 = − 2 k . Thus, v 1 = k and v 2 = − 2 k , the general solution to this system is, v = [ v 1 , v 2 ] T = [ k , − 2 k ] T = [ k − 2 k ] = k [ 1 − 2 ] : k ∈ ℝ Here, k is a scalar. The eigenvector corresponding to the eigenvalue λ 2 = − 4 is determined based on the linear system given as, ( A − λ I ) v = 0 ( [ − 12 − 7 16 10 ] − [ λ 0 0 λ ] ) [ v 1 v 2 ] = [ 0 0 ] [ − 12 − λ − 7 16 10 − λ ] [ v 1 v 2 ] = [ 0 0 ] Substitute − 4 for λ in the above equation

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Ch. 7.1 - Prob. 1P Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - Prob. 4P Ch. 7.1 - Prob. 5P Ch. 7.1 - Given that v1=(2,1) and v2=(1,1) are eigenvectors...Ch. 7.1 - Prob. 7P Ch. 7.1 - Prob. 8P Ch. 7.1 - Prob. 9P Ch. 7.1 - Prob. 11P

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