The approximated value of y ( 0.2 ) for initial value problem y ″ − 2 y ′ + 2 y = e t cos t , y ( 0 ) = 1 and y ′ ( 0 ) = 2 using step size h = 0.1 and h = 0.2 .

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Answer 1

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Chapter 9.4, Problem 5E

To determine

The approximated value of y(0.2) for initial value problem y″−2y′+2y=etcost, y(0)=1 and y′(0)=2 using step size h=0.1 and h=0.2.

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Answer 2

Question

Chapter 9.4, Problem 5E

To determine

The approximated value of y(0.2) for initial value problem y″−2y′+2y=etcost, y(0)=1 and y′(0)=2 using step size h=0.1 and h=0.2.

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Answer 3

Question

Chapter 9.4, Problem 5E

To determine

The approximated value of y(0.2) for initial value problem y″−2y′+2y=etcost, y(0)=1 and y′(0)=2 using step size h=0.1 and h=0.2.

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Answer 4

Question

Chapter 9.4, Problem 5E

To determine

The approximated value of y(0.2) for initial value problem y″−2y′+2y=etcost, y(0)=1 and y′(0)=2 using step size h=0.1 and h=0.2.

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Answer 5

Chapter 9.4, Problem 5E

To determine

The approximated value of y(0.2) for initial value problem y″−2y′+2y=etcost, y(0)=1 and y′(0)=2 using step size h=0.1 and h=0.2.

Answer 6

Chapter 9.4, Problem 5E

To determine

The approximated value of y(0.2) for initial value problem y″−2y′+2y=etcost, y(0)=1 and y′(0)=2 using step size h=0.1 and h=0.2.

Answer 7

Chapter 9.4, Problem 5E

To determine

The approximated value of y(0.2) for initial value problem y″−2y′+2y=etcost, y(0)=1 and y′(0)=2 using step size h=0.1 and h=0.2.

Answer 8

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Answer 9

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Answer 10

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Answer 11

The approximated value of y ( 0.2 ) for initial value problem y ″ − 2 y ′ + 2 y = e t cos t , y ( 0 ) = 1 and y ′ ( 0 ) = 2 using step size h = 0.1 and h = 0.2 .

Solution Summary: The author compares the approximated value of y(0.2) for the initial value problem and the second order differential equation.

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The approximated value of y(0.2) for initial value problem y″−2y′+2y=etcost, y(0)=1 and y′(0)=2 using step size h=0.1 and h=0.2.

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