Explanation Formula used: The solution to the initial value problem X ′ = AX, X ( 0 ) = X 0 is X ( t ) = Φ ( t ) Φ − 1 ( 0 ) X 0 . Here, Φ ( t ) is the fundamental matrix for the linear homogeneous system X ′ = AX . Find the eigenvalues by using the formula: | A − λI | = 0 A solution λ to this equation gives eigenvalues and vector v satisfying ( A − λI ) v = 0 is an eigenvectors corresponds to λ . Repeated eigenvalues: If A has repeated eigenvalues λ 1 = λ 2 with only one corresponding eigenvectors v, two linear independent solutions of X ′ = AX are X 1 = v 1 e λ 1 t and X 2 = ( v 1 t + w 2 ) e λ 1 t The formula for find the value of w 2 : ( A − λI ) w 2 = v 1 Inverse of the matrix is given by, A − 1 = A a | A | Here, A a is the transpose of the co factor matrix. Calculation: Consider the expression X ′ = ( 0 − 4 4 0 ) X . The eigenvalues of matrix A = ( 0 − 4 4 0 ) are obtained as follows. Apply the formula to find the eigenvalues. | A − λI | = | ( 0 − 4 4 0 ) − λ ( 1 0 0 1 ) | = | ( 0 − 4 4 0 ) − ( λ 0 0 λ ) | = | ( − λ − 4 4 − λ ) | = ( − λ ) ( − λ ) + 16 Further simplify as, ( − λ ) ( − λ ) + 16 = λ 2 + 16 Equate this to zero and find λ . λ 2 + 16 = 0 λ 2 = − 16 λ = ± 4 i The eigenvalues are λ 1 , 2 = ± 4 i . The eigenvector corresponds to λ 1 = 4 i is, ( A − λI ) = ( 0 − 4 4 0 ) − ( 4 i ) ( 1 0 0 1 ) = ( 0 − 4 4 0 ) − ( 4 i 0 0 4 i ) = ( − 4 i − 4 4 − 4 i ) The augmented matrix reduces as, ( − 4 i − 4 4 − 4 i ) → R 2 − i R 1 ( − 4 i − 4 0 0 ) It indicates that − 4 i x 1 − 4 y 1 = 0 . Choose x 1 = 1 , then y 1 = − i gives v 1 = ( 1 − i ) = ( 1 0 ) + ( 0 − 1 ) i . The eigenvalues λ 1 , 2 are complex conjugate. Note that α = 0 , β = 4 . Then, the value of a , b are obtained as follows. The value of a = ( 1 0 ) and b = ( 0 − 1 ) . Obtain X 1 as follows. X 1 = e − 0 t ( ( 1 0 ) cos 4 t − ( 0 − 1 ) sin 4 t ) = ( ( cos 4 t 0 ) − ( 0 − sin 4 t ) ) = ( cos 4 t sin 4 t ) Obtain X 2 as follows

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ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 6.4, Problem 39E

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Chapter 6.4, Problem 38E

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Introductory Differential Equations

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The solution of initial value problem X ′ = ( 0 − 4 4 0 ) X , X ( 0 ) = ( 0 8 ) .

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