Explanation Given: The given non-homogeneous system is, X ′ = ( 1 8 1 − 1 ) X + ( e − t t e 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 | 1 − λ 8 1 − 1 − λ | = 0 ( 1 − λ ) ( − 1 − λ ) − 8 = 0 − ( 1 − λ 2 ) − 8 = 0 Simplifying the above obtained expression gives, − 1 + λ 2 − 8 = 0 λ 2 − 9 = 0 λ = 9 λ = ± 3 So, Eigen values are λ 1 = 3 , λ 2 = − 3 , . Now evaluating the Eigen vector for λ 1 = 3 , ( A − 3 I 0 ) = ( − 2 8 0 1 − 4 0 ) On solving, it gives − 2 K 1 + 8 K 2 = 0 Which on further evaluating gives K 1 = 4 K 2 . Thus, the corresponding Eigen vector is K = ( 4 1 ) . Hence, the value of corresponding solution vector is X 1 = ( 4 1 ) e 3 t . Now evaluating the Eigen vector for λ 2 = − 3 , ( A − ( − 3 ) I 0 ) = ( 4 8 0 1 2 0 ) On solving, it gives 4 K 1 + 8 K 2 = 0 Which on further evaluating gives K 1 = − 2 K 2 . Thus, the corresponding Eigen vector is K = ( − 2 1 ) . Hence, the value of corresponding solution vector is X 2 = ( − 2 1 ) e − 3 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 = X 1 + X 2 Substituting the value of X 1 and X 2 in above expression it gives, X c = ( 4 1 ) e 3 t + ( − 2 1 ) e − 3 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 ) = ( 4 e 3 t − 2 e − 3 t e 3 t e − 3 t ) Evaluating determinant of the above matrix ϕ ( t ) , det [ ϕ ( t ) ] = | 4 e 3 t − 2 e − 3 t e 3 t e − 3 t | = 4 e 3 t ⋅ e − 3 t − ( − 2 e − 3 t ) e 3 t = 4 + 2 = 6 Now, the value of the value of ϕ − 1 ( t ) is ϕ − 1 ( t ) = 1 | ϕ ( t ) | ⋅ ( Adj

Bundle: A First Course in Differential Equations with Modeling Applications, Loose-leaf Version, 11th + WebAssign Printed Access Card for Zill's ... Problems, 9th Edition, Single-Term

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ISBN: 9781337761000

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 8.3, Problem 20E

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

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Chapter 8.3, Problem 19E

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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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