The blank in the statement: “The vector X = c 1 ( − 1 1 ) e − 9 t + c 2 ( 5 3 ) e 7 t is a solution of the initial-value problem X ′ = ( 1 10 6 − 3 ) X , X ( 0 ) = ( 2 0 ) for c 1 = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ and c 2 = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .”

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Answer 1

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Chapter 10, Problem 2CR

To determine

The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”

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Linear Algebra 2. Find all the values of unknown variables which satisfy the following equation: (z4, z2, 1) (1, 1, -z2) = tan(π/4) Find all the values of z.

Answer 2

Question

Chapter 10, Problem 2CR

To determine

The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”

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Answer 3

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Chapter 10, Problem 2CR

To determine

The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”

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Answer 4

Question

Chapter 10, Problem 2CR

To determine

The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”

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Answer 5

Chapter 10, Problem 2CR

To determine

The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”

Answer 6

Chapter 10, Problem 2CR

To determine

The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”

Answer 7

Chapter 10, Problem 2CR

To determine

The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

The blank in the statement: “The vector X = c 1 ( − 1 1 ) e − 9 t + c 2 ( 5 3 ) e 7 t is a solution of the initial-value problem X ′ = ( 1 10 6 − 3 ) X , X ( 0 ) = ( 2 0 ) for c 1 = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ and c 2 = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .”

Solution Summary: The author explains that the vector X=c_1(

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The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”

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Chapter 10 Solutions

The blank in the statement: “The vector X = c 1 ( − 1 1 ) e − 9 t + c 2 ( 5 3 ) e 7 t is a solution of the initial-value problem X ′ = ( 1 10 6 − 3 ) X , X ( 0 ) = ( 2 0 ) for c 1 = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ and c 2 = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .”

Solution Summary: The author explains that the vector X=c_1(

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The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_________________ and c2=_________________.”

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Chapter 10 Solutions

The blank in the statement: “The vector X=c1(−1 1)e−9t+c2(53)e7t is a solution of the initial-value problem X′=(1 106−3)X, X(0)=(20) for c1=_ and c2=_.”