Explanation Given: The given system is: X ′ ( t ) = A X ( t ) + f ( t ) Here, A = [ 1 2 3 2 ] , f ( t ) = [ 1 − 1 ] Procedure used: Variation of parameters: Suppose n × n constant matrix A has n linearly independent eigenvectors u 1 , u 2 , ⋯ , u n . Let r i be the eigenvalues corresponding to u i . Then, { e r 1 t u 1 , e r 2 t u 2 , ⋯ , e r n t u n } is a fundamental solution set and [ e r 1 t u 1 , e r 2 t u 2 , ⋯ , e r n t u n ] is called fundamental matrix. Let X ( t ) is the fundamental matrix for the non-homogeneous system X ′ ( t ) = A ( t ) X ( t ) + f ( t ) . Hence, the particular solution of non-homogeneous system is X p ( t ) = X ( t ) v ( t ) = X ( t ) ∫ X − 1 ( t ) f ( t ) d t . The general solution is the combination of X ( t ) c and X p ( t ) . x ( t ) = X ( t ) c + X p ( t ) Calculation: Consider the given matrix: A = [ 1 2 3 2 ] Substitute the value of matrix A in the characteristic equation and solve for r : | [ 1 2 3 2 ] − r [ 1 0 0 1 ] | = 0 | [ 1 2 3 2 ] − [ r 0 0 r ] | = 0 | [ 1 − r 2 3 2 − r ] | = 0 ( 1 − r ) ( 2 − r ) − 6 = 0 Further simplify as: 2 − r − 2 r + r 2 − 6 = 0 r 2 − 3 r − 4 = 0 r 2 − 4 r + r − 4 = 0 r ( r − 4 ) + 1 ( r − 4 ) = 0 Further simplify as: ( r − 4 ) ( r + 1 ) = 0 Equate first factor to 0: ( r − 4 ) = 0 r = 4 Equate second factor to 0: ( r + 1 ) = 0 r = − 1 Therefore, the eigenvalues are r 1 = − 1 and r 2 = 4 . Find the eigenvector corresponding to eigenvalue r 1 = − 1 by using the formula ( A − r I ) u = 0 : ( A − r I ) u = ( [ 1 2 3 2 ] − ( − 1 ) [ 1 0 0 1 ] ) [ u 1 u 2 ] = ( [ 1 2 3 2 ] + [ 1 0 0 1 ] ) [ u 1 u 2 ] = [ 2 2 3 3 ] [ u 1 u 2 ] The matrix equation is: [ 2 2 3 3 ] [ u 1 u 2 ] = [ 0 0 ] It can be write in scalar equation as, u 1 + u 2 = 0 . Choose the arbitrary value of u 2 = 1 , then u 1 = − 1 . The eigenvector can be expressed as: u 1 = [ − 1 1 ] Find the eigenvector corresponding to eigenvalue r 2 = 4 by using the formula ( A − r I ) u = 0 : ( A − r I ) u = ( [ 1 2 3 2 ] − 4 [ 1 0 0 1 ] ) [ u 1 u 2 ] = ( [ 1 2 3 2 ] − [ 4 0 0 4 ] ) [ u 1 u 2 ] = [ − 3 2 3 − 2 ] [ u 1 u 2 ] The matrix equation is: [ − 3 2 3 − 2 ] [ u 1 u 2 ] = [ 0 0 ] It can be write in scalar equation as, − 3 u 1 + 2 u 2 = 0

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Chapter 9.7, Problem 12E

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The general solution by using variation of parameters formula for the given system.

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To find: The general solution by using variation of parameters formula for the given system.

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Chapter 9 Solutions

To find:

The general solution by using variation of parameters formula for the given system.