Explanation The given non-homogeneous system is, X ′ = ( 3 2 − 2 − 1 ) X + ( 1 1 ) 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 | 3 − λ 2 − 2 − 1 − λ | = 0 ( 3 − λ ) ( − 1 − λ ) + 4 = 0 − 3 − 3 λ + λ + λ 2 + 4 = 0 Simplifying the above obtained expression gives, λ 2 − 2 λ + 1 = 0 ( λ − 1 ) 2 = 0 λ = 1 , 1 So, Eigen values are λ 1 = 1 , λ 2 = 1 , . Now evaluating the Eigen vector for λ 1 = 1 , ( A − I 0 ) = ( 2 2 0 − 2 − 2 0 ) On solving, it gives 2 K 1 + 2 K 2 = 0 Which on further evaluating gives K 1 = − K 2 . Thus, the corresponding Eigen vector is K = ( 1 − 1 ) . Hence, the value of corresponding solution vector is X 1 = ( 1 − 1 ) e t . Since, there are two repeated Eigen value, so consider another Eigen vector as P = ( a b ) . Now, solve the next system ( A − 2 I ) P = K to get the value of another Eigen vector. ( A − 2 I ) ( a b ) = K ( 2 2 − 2 − 2 ) ( a b ) = ( 1 − 1 ) Apply the operation R 1 → R 1 + R 2 in the above expression. ( 2 2 0 0 ) ( a b ) = ( 1 0 ) On solving, it gives 2 a + 2 b = 1 In the above obtained expression, there is only one equation and two variables, so consider a = 0 , then b becomes 1 2 . Thus, the corresponding Eigen vector is P = ( 0 1 2 ) . Hence, the value of corresponding solution vector is, X 2 = K t e λ 1 t + P e λ 2 t = ( 1 − 1 ) t e t + ( 0 1 2 ) e 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 − 1 ) t e t + C 2 [ ( 1 − 1 ) t e t + ( 0 1 2 ) e 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 t t e t − e t 1 2 e t − t e t ) Evaluating determinant of the above matrix ϕ ( t ) , det [ ϕ ( t ) ] = | e t t e t − e t 1 2 e t − t e t | = e t ( 1 2 e t − t e t ) − t e t ( − e t ) = 1 2 e 2 t − t e 2 t + t e 2 t = 1 2 e 2 t Now, the value of the value of ϕ − 1 ( t ) is ϕ − 1 ( t ) = 1 | ϕ ( 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 22E

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

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Chapter 8.3, Problem 21E

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

Solution Summary: The author explains how to solve the given non-homogeneous system by using variation of parameters. The characteristic equation of a matrix A is left|A-lambda Iright|=

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