Explanation Given: The linear system is X ′ = ( 3 1 − 1 1 ) X + ( − 2 1 ) e 2 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 ) = ( 3 − λ 1 − 1 1 − λ ) For evaluation of the values of λ , the determinant of the coefficient matrix equate to zero. | ( A − λ I ) | = 0 | 3 − λ 1 − 1 1 − λ | = 0 ( 3 − λ ) ( 1 − λ ) + 1 = 0 3 − 3 λ − λ + λ 2 + 1 = 0 Further simplify the above equation to obtain the values of λ . λ 2 − 4 λ + 4 = 0 ( λ − 2 ) 2 = 0 λ 1 = 2 , λ 2 = 2 For λ 1 = 2 , the Eigen vector is given below. ( A − 2 I 0 ) = ( 1 1 0 − 1 − 1 0 ) Therefore, the value of K 1 + K 2 = 0 . From the above expression K 1 = − K 2 . The Corresponding Eigen vector is K = ( 1 − 1 ) . For two repeated Eigen value, consider a vector P = ( a b ) . The system ( A − 2 I ) P = K is calculated as follows: ( A − 2 I ) ( a b ) = ( 1 − 1 ) ( 1 1 − 1 − 1 ) ( a b ) = ( 1 − 1 ) Therefore, the value of a + b = 1 . There is only one equation and two variables, so consider a = 0 then b becomes 1 . Here, K 1 , K 2 are the components of the Eigen vector. The Corresponding Eigen vector is P = ( 1 0 ) . The solution of the given system is X = C 1 ( 1 − 1 ) e 2 t + C 2 [ ( 1 − 1 ) t e 2 t + ( 1 0 ) e 2 t ] . The solution system of the given system can be expressed as X 1 = ( 1 − 1 ) e 2 t , X 2 = ( 1 − 1 ) t e 2 t + ( 1 0 ) e 2 t . The solution vector can be written in the simplest form of the vectors X 1 and X 2 . The solution vectors can be expressed as given below. ϕ ( t ) = ( e 2 t t e 2 t + e 2 t − e 2 t − t e 2 t ) The determinant of the solution vector can be evaluated as follows: | ϕ ( t ) | = | e 2 t t e 2 t + e 2 t − e 2 t − t e 2 t | = − t e 4 t − ( − t e 4 t − e 4 t ) = e 4 t The inverse of the solution vector can be evaluated as follows: ϕ − 1 ( t ) = 1 ϕ − 1 ( t ) ( − t e 2 t − ( t e 2 t + e 2 t ) e 2 t e 2 t ) = ( − t e − 2 t − t e − 2 t − e − 2 t e − 2 t e − 2 t ) The particular solution X p can be expressed as given below

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Author: ZILL

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Chapter 8, Problem 14RE

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The solution of the linear system X′=(31−11)X+(−21)e2t.

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The solution of the linear system X ′ = ( 3 1 − 1 1 ) X + ( − 2 1 ) e 2 t .

Solution Summary: The author explains the solution of the linear system Xprime and the coefficient matrix A.

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The solution of the linear system X′=(31−11)X+(−21)e2t.

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