Explanation Given: The given non-homogeneous system is, X ′ = ( 1 1 0 1 1 0 0 0 3 ) X + ( e t e 2 t t e 3 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 1 0 1 1 − λ 0 0 0 3 − λ | = 0 ( 3 − λ ) [ ( 1 − λ ) 2 − 1 ] = 0 Simplify the above obtained expression, ( 3 − λ ) [ λ 2 − 2 λ + 1 − 1 ] = 0 λ ( 3 − λ ) ( λ − 2 ) = 0 λ = 0 , 2 , 3 So, Eigen values are λ 1 = 0 , λ 2 = 2 , λ 3 = 3 . Now evaluating the Eigen vector for λ 1 = 0 , ( A − ( 0 ) I 0 ) = ( 1 1 0 0 1 1 0 0 0 0 3 0 ) On solving, it gives K 1 + K 2 = 0 3 K 3 = 0 Which on further evaluation becomes K 1 = − K 2 and K 3 = 0 . So the corresponding Eigen vector is K = ( 1 − 1 0 ) . Thus, the value of corresponding solution vector is X 1 = ( 1 − 1 0 ) . Now evaluating the Eigen vector for λ 2 = 2 , ( A − 2 I 0 ) = ( − 1 1 0 0 1 − 1 0 0 0 0 1 0 ) On solving, it gives − K 1 + K 2 = 0 K 3 = 0 Which, on further evaluation becomes K 1 = K 2 . So the corresponding Eigen vector is K = ( 1 1 0 ) . Thus, the value of corresponding solution vector is X 2 = ( 1 1 0 ) e 2 t . Now evaluating the Eigen vector for λ 3 = 3 , ( A − 3 I 0 ) = ( − 2 1 0 0 1 − 2 0 0 0 0 0 0 ) On solving, it gives − 2 K 1 + K 2 = 0 which on further evaluation becomes 2 K 1 = K 2 . Since, there is no equation involving K 3 , so choose an arbitrary value of K 3 as K 3 = 1 and assume K 2 = 0 which implies K 1 = 0 So the corresponding Eigen vector is K = ( 0 0 1 ) . Thus, the value of corresponding solution vector is X 3 = ( 0 0 1 ) e 3 t . Consider the coefficient matrix A having real Eigen values of homogeneous system and K be an Eigen vector and X 1 , X 2 and X 3 be the solution vectors for the corresponding Eigen value. Then, the value of complementary function X c is X c = C 1 X 1 + C 2 X 2 + C 3 X 3 Substituting the value of X 1 , X 2 and X 3 in above expression it gives, X c = C 1 ( 1 − 1 0 ) + C 2 ( 1 1 0 ) e 2 t + C 3 ( 0 0 1 ) e 3 t Now, the entries in X 1 form the first column of solution matrix ϕ ( t ) , the entries in X 2 form the second column, the entries in X 3 form the third column of ϕ ( t ) , which means the value of ϕ ( t ) is ϕ ( t ) = ( 1 e 2 t 0 − 1 e 2 t 0 0 0 e 3 t ) Now, evaluating determinant of the above matrix ϕ ( t ) gives det [ ϕ ( t ) ] = | 1 e 2 t 0 − 1 e 2 t 0 0 0 e 3 t | = e 3 t ( e 2 t + e 2 t ) = 2 e 5 t Now, evaluating co-factor matrix of the above matrix ϕ ( t ) gives C = ( e 5 t e 3 t 0 e 5 t e 3 t 0 0 0 2 e 2 t ) Now, the transpose of the co-factor matrix gives Adj ϕ ( t )

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 31E

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