Explanation Given: The given system of differential equations, x ′ 1 = x 1 + x 2 − x 3 , x ′ 2 = x 1 + x 2 + x 3 and x ′ 3 = − x 1 + x 2 + x 3 . Approach: Let, the differential equations are, x ′ 1 = a 1 x 1 + b 1 x 2 + c 1 x 3 x ′ 2 = a 2 x 1 + b 2 x 2 + c 2 x 3 x ′ 3 = a 3 x 1 + b 3 x 2 + c 3 x 3 The given system is written as, x ' = A x Here, matrix A is, A = [ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ] 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 = x 1 + x 2 − x 3 x ′ 2 = x 1 + x 2 + x 3 x ′ 3 = − x 1 + x 2 + x 3 The given system is written as, x ′ = A x [ x ′ 1 x ′ 2 x ′ 3 ] = [ 1 1 − 1 1 1 1 − 1 1 1 ] [ x 1 x 2 x 3 ] Here, matrix A is, A = [ 1 1 − 1 1 1 1 − 1 1 1 ] Consider the matrix A . A = [ 1 1 − 1 1 1 1 − 1 1 1 ] The characteristic equation of A is given as det ( A − λ I ) = 0 . det ( A − λ I ) = 0 | [ 1 1 − 1 1 1 1 − 1 1 1 ] − [ λ 0 0 0 λ 0 0 0 λ ] | = 0 | [ 1 − λ 1 − 1 1 1 − λ 1 − 1 1 1 − λ ] | = 0 Solve the above equation for the eigenvalues ( λ ) of A . | 1 − λ 1 − 1 1 1 − λ 1 − 1 1 1 − λ | = 0 ( 1 − λ ) [ ( 1 − λ ) ( 1 − λ ) − ( 1 ) ( 1 ) ] − ( 1 ) [ ( 1 ) ( 1 − λ ) − ( − 1 ) ( 1 ) ] + ( − 1 ) [ ( 1 ) ( 1 ) − ( − 1 ) ( 1 − λ ) ] = 0 ( 1 − λ ) [ ( 1 − λ ) 2 − 1 ] − [ ( 1 − λ ) + 1 ] − [ 1 + ( 1 − λ ) ] = 0 ( 1 − λ ) [ ( 1 − λ ) 2 − 1 ] − 2 [ ( 1 − λ ) + 1 ] = 0 Further solve the above equation. ( 1 − λ ) 3 − ( 1 − λ ) − 2 ( 2 − λ ) = 0 ( 1 − λ ) 3 − 1 + λ + 2 λ − 4 = 0 ( 1 − λ ) 3 + 3 λ − 5 = 0 − λ 3 + 3 λ 2 − 3 λ + 1 + 3 λ − 5 = 0 Further solve the above equation. − λ 3 + 3 λ 2 − 4 = 0 − ( λ − 2 ) 2 ( λ + 1 ) = 0 ( λ − 2 ) 2 ( λ + 1 ) = 0 λ = − 1 , 2 , 2 Therefore, the eigenvalue of A are − 1 , 2 and 2 . The eigenvector corresponding to the eigenvalue λ 1 = − 1 is determined based on the linear system given as, ( A − λ I ) v = 0 ( [ 1 1 − 1 1 1 1 − 1 1 1 ] − [ λ 0 0 0 λ 0 0 0 λ ] ) [ v 1 v 2 v 3 ] = [ 0 0 0 ] [ 1 − λ 1 − 1 1 1 − λ 1 − 1 1 1 − λ ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] Substitute − 1 for λ in the above equation. [ 1 − ( − 1 ) 1 − 1 1 1 − ( − 1 ) 1 − 1 1 1 − ( − 1 ) ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] [ 1 + 1 1 − 1 1 1 + 1 1 − 1 1 1 + 1 ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] [ 2 1 − 1 1 2 1 − 1 1 2 ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] The augmented matrix of the above expression is, [ 2 1 − 1 1 2 1 − 1 1 2 | 0 0 0 ] Apply R 2 → R 2 − R 1 and R 3 → 2 R 3 − R 1 . [ 2 1 − 1 0 3 3 0 3 3 | 0 0 0 ] Apply R 3 → R 3 − R 2 . [ 2 1 − 1 0 3 3 0 0 0 | 0 0 0 ] Apply R 2 → 1 3 R 2 . [ 2 1 − 1 0 1 1 0 0 0 | 0 0 0 ] Apply R 1 → R 1 + R 2 . [ 2 2 0 0 1 1 0 0 0 | 0 0 0 ] Apply R 1 → 1 2 R 1 . [ 1 1 0 0 1 1 0 0 0 | 0 0 0 ] Thus, [ 1 1 0 0 1 1 0 0 0 ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] [ v 1 + v 2 v 2 + v 3 0 ] = [ 0 0 0 ] The obtained equation is, v 1 + v 2 = 0 v 1 = − v 2 And, v 2 + v 3 = 0 v 2 = − v 3 Let, v 2 = − k , then v 1 = k and v 3 = k . Thus, v 1 = k , v 2 = − k and v 3 = k , the general solution to this system is, v = [ v 1 , v 2 , v 3 ] T = [ k , − k , k ] T = [ k − k k ] = k [ 1 − 1 1 ] : k ∈ ℝ Here, k is a scalar

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Chapter 7.3, Problem 24P

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Chapter 7 Solutions

EBK DIFFERENTIAL EQUATIONS AND LINEAR A

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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The solution of the given system of differential equations based on the idea introduced in this section.

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Chapter 7 Solutions