Explanation Given: The linear system is X ′ = ( 1 2 − 1 2 1 ) X + ( 0 e t tan t ) . Calculation: The system is in the form X ′ = A X + F ( t ) . Write the characteristic equation of the coefficient matrix A . ( A − λ I ) = ( 1 − λ 2 − 1 2 1 − λ ) For the Eigen values, determinant of the coefficient matrix equate to zero. Det ( A − λ I ) = | 1 − λ 2 − 1 2 1 − λ | ( 2 − λ ) 2 + 1 = 0 λ 2 − 2 λ + 2 = 0 The values of λ is calculated as follows: λ = 2 ± 4 − 4 ⋅ 2 2 λ = 1 ± i λ 1 = 1 + i or λ 2 = 1 − i where λ 1 , λ 2 are the two Eigen values. For λ 1 = 1 + i , the Eigen vector is given below. ( A − ( 1 + i ) I | 0 ) = ( − i 2 0 − 1 2 − i 0 ) Therefore, the value is − i K 1 + 2 K 2 = 0 From the above expression K 2 = i 2 K 1 . The Corresponding Eigen vector is K 1 = ( 2 i ) . Consider A is the coefficient matrix having real values of homogeneous system and K is an Eigen vector corresponding the complex Eigen value λ 1 = 1 + i then the solution is X = K e λ 1 t and this solution will come in the form of X = X 1 + X 2 . X = ( 2 i ) e ( 1 + i ) t = ( 2 i ) e t ( cos t + i sin t ) = ( 2 cos t − sin t ) e t + i ( 2 sin t cos t ) e t The solution vector of the system is given below. X 1 = ( 2 cos t − sin t ) e t , X 2 = ( 2 sin t cos t ) e t The solution vectors can be written as follows: ϕ ( t ) = ( 2 cos t ⋅ e t 2 sin t ⋅ e t − sin t ⋅ e t cos t ⋅ e t ) Det ϕ ( t ) = ( 2 cos t ⋅ e t ) ⋅ ( cos t ⋅ e t ) − ( − sin t ⋅ e t ) ⋅ ( cos t ⋅ e t ) = 2 e t cos 2 t + 2 e 2 t sin 2 t = 2 e 2 t Determination of ϕ − 1 ( t ) is calculated as follows: ϕ − 1 ( t ) = 1 ϕ − 1 ( t ) ( cos t ⋅ e t − 2 sin t ⋅ e t sin t ⋅ e t 2 cos t ⋅ e t ) ϕ − 1 ( t ) = ( 1 2 e − t cos t − e − t sin t 1 2 e − t sin t ⋅ e − t cos t ) Express the particular solution X p as given below. X p = ϕ ( t ) ∫ ϕ − 1 ( t ) F ( t ) d t

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 8, Problem 12RE

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The solution of the linear system X′=(12−121)X+(0ettant).

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The solution of the linear system X ′ = ( 1 2 − 1 2 1 ) X + ( 0 e t tan t ) .

Solution Summary: The author explains the linear system Xprime, which is underset_x=1left.

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The solution of the linear system X′=(12−121)X+(0ettant).

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