Explanation Given: The given non-homogeneous system is, X ′ = ( 2 − 2 8 − 6 ) X + ( 1 3 ) e − 2 t t Calculation: The system is given in the form X ′ = A X + F ( t ) (1). Consider, the characteristic equation of a matrix A as | A − λ I | = 0 , where I is identity matrix. Now, the characteristic equation of the coefficient matrix A . is | ( A − λ I ) | = 0 | 2 − λ − 2 8 − 6 − λ | = 0 ( 2 − λ ) ( − 6 − λ ) + 16 = 0 − 12 − 2 λ + 6 λ + λ 2 + 16 = 0 Simplifying the above obtained expression it gives, λ 2 + 4 λ + 4 = 0 ( λ + 2 ) 2 = 0 λ = − 2 , − 2 So, Eigen values are λ 1 = − 2 , λ 2 = − 2 , . Now evaluating the Eigen vector for λ 1 = − 2 , ( A − ( − 2 ) I 0 ) = ( 4 − 2 0 8 − 4 0 ) On solving, it gives 4 K 1 − 2 K 2 = 0 which on further evaluation gives 2 K 1 = K 2 . So, the corresponding Eigen vector is K = ( 1 2 ) . Thus, the value of corresponding solution vector is X 1 = ( 1 − 1 ) e t . There are two repeated Eigen value, so consider another Eigen vector is P = ( a b ) . Now, solve the next system ( A − 2 I ) P = K to get the another Eigen vector. ( A − ( − 2 ) I ) ( a b ) = K ( 4 − 2 8 − 4 ) ( a b ) = ( 1 2 ) On solving, it gives 4 a − 2 b = 1 In the above obtained expression, there is only one equation and two variables, so for satisfying the above expression consider a = 1 2 , then b becomes 1 2 . So the corresponding Eigen vector is P = ( 1 2 1 2 ) . Thus, the value of corresponding solution vector X 2 is, X 2 = K t e λ 1 t + P e λ 2 t = ( 1 2 ) t e − 2 t + ( 1 2 1 2 ) e − 2 t . Consider the coefficient matrix A having real Eigen values of homogeneous system, K be an Eigen vector and X 1 , X 2 be the solution vectors of the corresponding Eigen values then, the value of complementary function X c is, X c = C 1 X 1 + C 2 X 2 Substituting the value of X 1 and X 2 in above expression it gives, X c = C 1 ( 1 2 ) e − 2 t + C 2 [ ( 1 2 ) t e − 2 t + ( 1 2 1 2 ) e − 2 t ] (2) Now, the entries of X 1 form the first column of solution matrix ϕ ( t ) , and the entries of X 2 form the second column of ϕ ( t ) , which means the value of ϕ ( t ) is ϕ ( t ) = ( e − 2 t ( t + 1 2 ) e − 2 t 2 e − 2 t ( 2 t + 1 2 ) e − 2 t ) Evaluating determinant of the above matrix ϕ ( t ) , det [ ϕ ( t ) ] = | e − 2 t ( t + 1 2 ) e − 2 t 2 e − 2 t ( 2 t + 1 2 ) e − 2 t | = e − 2 t ⋅ ( 2 t + 1 2 ) e − 2 t − 2 e − 2 t ⋅ ( t + 1 2 ) e − 2 t = 2 t e − 4 t + 1 2 e − 4 t − 2 t e − 4 t − e − 4 t = − 1 2 e − 4 t Now, the value of the value of ϕ − 1 ( t ) is ϕ − 1 ( t ) = 1 | ϕ ( t ) | ⋅ ( Adj . ϕ ( t ) ) = 1 ( − 1 2 e − 4 t t ) ⋅ [ Adj

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 8.3, Problem 26E

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To solve: The given non-homogeneous system by using variation of parameters.

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