The analytic solution of the initial value problem y ′ = x + y − 1 , y ( 0 ) = 1 and compare the actual values y ( 0.2 ) , y ( 0.4 ) , y ( 0.6 ) and y ( 0.8 ) with approximate values in example 1.

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Answer 1

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Chapter 6.3, Problem 1E

To determine

The analytic solution of the initial value problem y′=x+y−1, y(0)=1 and compare the actual values y(0.2), y(0.4), y(0.6) and y(0.8) with approximate values in example 1.

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Answer 2

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Chapter 6.3, Problem 1E

To determine

The analytic solution of the initial value problem y′=x+y−1, y(0)=1 and compare the actual values y(0.2), y(0.4), y(0.6) and y(0.8) with approximate values in example 1.

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Answer 3

Question

Chapter 6.3, Problem 1E

To determine

The analytic solution of the initial value problem y′=x+y−1, y(0)=1 and compare the actual values y(0.2), y(0.4), y(0.6) and y(0.8) with approximate values in example 1.

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Answer 4

Question

Chapter 6.3, Problem 1E

To determine

The analytic solution of the initial value problem y′=x+y−1, y(0)=1 and compare the actual values y(0.2), y(0.4), y(0.6) and y(0.8) with approximate values in example 1.

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Answer 5

Chapter 6.3, Problem 1E

To determine

The analytic solution of the initial value problem y′=x+y−1, y(0)=1 and compare the actual values y(0.2), y(0.4), y(0.6) and y(0.8) with approximate values in example 1.

Answer 6

Chapter 6.3, Problem 1E

To determine

The analytic solution of the initial value problem y′=x+y−1, y(0)=1 and compare the actual values y(0.2), y(0.4), y(0.6) and y(0.8) with approximate values in example 1.

Answer 7

Chapter 6.3, Problem 1E

To determine

The analytic solution of the initial value problem y′=x+y−1, y(0)=1 and compare the actual values y(0.2), y(0.4), y(0.6) and y(0.8) with approximate values in example 1.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - Prob. 7E Ch. 6.1 - Prob. 8E Ch. 6.1 - Prob. 9E Ch. 6.1 - Prob. 10E

The analytic solution of the initial value problem y ′ = x + y − 1 , y ( 0 ) = 1 and compare the actual values y ( 0.2 ) , y ( 0.4 ) , y ( 0.6 ) and y ( 0.8 ) with approximate values in example 1.

Solution Summary: The author explains the analytic solution of the initial value problem yprime. The linear differential equation is rewritten as follows.

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The analytic solution of the initial value problem y′=x+y−1, y(0)=1 and compare the actual values y(0.2), y(0.4), y(0.6) and y(0.8) with approximate values in example 1.

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Chapter 6 Solutions