Explanation Given: The homogeneous linear system is X ′ = ( 0 6 0 1 0 1 1 1 0 ) X . The general solution of the above homogeneous linear system is X = c 1 ( 6 − 1 − 5 ) e − t + c 2 ( − 3 1 1 ) e − 2 t + c 3 ( 2 1 1 ) e 3 t . Calculation: Consider the given system in X ′ = A X form, where A = ( 0 6 0 1 0 1 1 1 0 ) . From the given general solution of the homogenous system, assume three solution vectors are as follows: X 1 = ( 6 − 1 − 5 ) e − t X 2 = ( − 3 1 1 ) e − 2 t X 3 = ( 2 1 1 ) e 3 t Differentiate each vector solution as follows: X ′ 1 = ( 6 − 1 − 5 ) ( − e − t ) X ′ 2 = ( − 3 1 1 ) ( − 2 e − 2 t ) X ′ 3 = ( 2 1 1 ) ( 3 e 3 t ) Simplify, each differentiate vector as follows: X ′ 1 = ( − 6 e − t e − t 5 e − t ) X ′ 2 = ( 6 e − 2 t − 2 e − 2 t − 2 e − 2 t ) X ′ 3 = ( 6 e 3 t 3 e 3 t 3 e 3 t ) Calculate the value of A X 1 as follows: A X 1 = ( 0 6 0 1 0 1 1 1 0 ) ( 6 − 1 − 5 ) e − t = ( − 6 1 5 ) e − t = ( − 6 e − t e − t 5 e − t ) = X ′ 1 From above calculation, it is proved that X 1 is the solution of homogenous system X ′ = A X . Calculate the value of A X 2 as follows: A X 2 = ( 0 6 0 1 0 1 1 1 0 ) ( − 3 1 1 ) e − 2 t = ( 6 − 2 − 2 ) e − 2 t = ( 6 e − 2 t − 2 e − 2 t − 2 e − 2 t ) = X ′ 2 From above calculation, it is proved that X 2 is the solution of homogenous system X ′ = A X

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Chapter 10.1, Problem 25E

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To prove: The general solution of the homogeneous linear system X′=(060101110)X on the interval (−∞,∞) is X=c1(6−1−5)e−t+c2(−311)e−2t+c3(211)e3t.

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Chapter 10.1, Problem 24E

Chapter 10.1, Problem 26E

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To prove: The general solution of the homogeneous linear system X ′ = ( 0 6 0 1 0 1 1 1 0 ) X on the interval ( − ∞ , ∞ ) is X = c 1 ( 6 − 1 − 5 ) e − t + c 2 ( − 3 1 1 ) e − 2 t + c 3 ( 2 1 1 ) e 3 t .

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To prove: The general solution of the homogeneous linear system X′=(060101110)X on the interval (−∞,∞) is X=c1(6−1−5)e−t+c2(−311)e−2t+c3(211)e3t.

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Chapter 10 Solutions