Explanation Consider the matrix, [ 2 − 1 3 − 2 ] ... ( 1 ) The characteristic equation of the matrix A is, | 2 − r − 1 3 − 2 − r | = 0 − ( 2 − r ) ( 2 + r ) + 3 = 0 r = − 1 , 1 The Eigen vector corresponding to r = 1 are, u 1 = [ 1 1 ] The Eigen vector corresponding to r = − 1 are, u 2 = [ 1 3 ] The solution of the system is, x ′ ( u ) = [ 2 − 1 3 − 2 ] x ( u ) Substitute [ e u e − u e u 3 e − u ] for x ( u ) in Equation (2). x ′ ( u ) = [ 2 − 1 3 − 2 ] [ e u e − u e u 3 e − u ] = [ e u − e − u e u − 3 e − u ] The inverse of matrix is, x − 1 ( u ) = 1 − 3 + 1 [ − 3 e − u e − u − e u e u ] = [ 3 2 e − u − 1 2 e − u 1 2 e u − 1 2 e u ] The general solution of system is, x ( u ) = x ( u ) C + x ( u ) ∫ x − 1 ( u ) f ( u ) d t

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Chapter 9.7, Problem 29E

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The general solution set for the system tx′(t)=[2−13−2]x(t)+[t−11].

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Chapter 9.7, Problem 28E

Chapter 9.7, Problem 30E

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The general solution set for the system t x ′ ( t ) = [ 2 − 1 3 − 2 ] x ( t ) + [ t − 1 1 ] .

Solution Summary: The author explains that the general solution set for the system is txprime (t-)= left[cc2

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The general solution set for the system tx′(t)=[2−13−2]x(t)+[t−11].

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Chapter 9 Solutions