Explanation Procedure used: Solving initial value problem: 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 to find the value of w 2 is, ( A − λI ) w 2 = v 1 Inverse of the matrix is, A − 1 = A a | A | Here, A a is the transpose of the co factor matrix. Calculation: Consider the expression X ′ = ( 0 13 − 1 − 4 ) X . The eigenvalues of matrix A = ( 0 13 − 1 − 4 ) are obtained as follows. Apply the formula to find the eigenvalues. | A − λI | = | ( 0 13 − 1 − 4 ) − λ ( 1 0 0 1 ) | = | ( 0 13 − 1 − 4 ) − ( λ 0 0 λ ) | = | ( − λ 13 − 1 − 4 − λ ) | = ( − λ ) ( − 4 − λ ) + 13 Further simplify as, ( − λ ) ( − 4 − λ ) + 13 = ( λ ) ( 4 + λ ) + 13 = λ 2 + 4 λ + 13 Equate to zero and find λ . λ 2 + 4 λ + 13 = 0 The above equation is a quadratic equation. Therefore, λ = − 4 ± 4 2 − 4 ( 13 ) 2 = − 4 ± 16 − 52 2 = − 4 ± − 36 2 = − 2 ± 3 i Therefore, the eigenvalues are λ 1 , 2 = − 2 ± 3 i . The eigenvector corresponds to λ 1 = − 2 + 3 i is, ( A − λI ) = ( 0 13 − 1 − 4 ) − ( − 2 + 3 i ) ( 1 0 0 1 ) = ( 0 13 − 1 − 4 ) − ( − 2 + 3 i 0 0 − 2 + 3 i ) = ( 2 − 3 i 13 − 1 − 2 − 3 i ) The augmented matrix reduces as, ( 2 − 3 i 13 − 1 − 2 − 3 i ) → 13 R 2 + ( 2 + 3 i ) R 1 ( 2 − 3 i 13 0 0 ) This indicates that, ( 2 − 3 i ) x 1 + 13 y 1 = 0 . Choose x 1 = 2 + 3 i , then y 1 = − 1 gives v 1 = ( 2 + 3 i − 1 ) = ( 2 − 1 ) + ( 3 0 ) i . The eigenvalues λ 1 , 2 are complex conjugates. Note that α = − 2 , β = 3 . Then, the values of a , b are obtained as follows. The value of a = ( 2 − 1 ) and b = ( 3 0 )

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Publisher: Elsevier Science

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Differential Equations

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Chapter 6.4, Problem 40E

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Chapter 6.4, Problem 39E

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Introductory Differential Equations

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The solution of initial value problem X ′ = ( 0 13 − 1 − 4 ) X, X ( 0 ) = ( 0 4 ) .

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The solution of initial value problem X′=(013−1−4)X, X(0)=(04).

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Chapter 6 Solutions