To prove: The initial value problem y ″ ( x ) + y ( x ) = 6 x y ( 0 ) = 1 , y ′ ( 0 ) = − 1 is equivalent to the integral equation y ( x ) = x 3 − x + 1 + ∫ 0 x ( u − x ) y ( u ) d u .

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Answer 1

Question

Chapter 13.2, Problem 10E

To determine

To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

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Answer 2

Question

Chapter 13.2, Problem 10E

To determine

To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

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Answer 3

Question

Chapter 13.2, Problem 10E

To determine

To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

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Answer 4

Question

Chapter 13.2, Problem 10E

To determine

To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

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Answer 5

Chapter 13.2, Problem 10E

To determine

To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

Answer 6

Chapter 13.2, Problem 10E

To determine

To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

Answer 7

Chapter 13.2, Problem 10E

To determine

To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 13.1 - In Problem 1-4, express the given initial value...Ch. 13.1 - In Problem 1-4, express the given initial value...Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

To prove: The initial value problem y ″ ( x ) + y ( x ) = 6 x y ( 0 ) = 1 , y ′ ( 0 ) = − 1 is equivalent to the integral equation y ( x ) = x 3 − x + 1 + ∫ 0 x ( u − x ) y ( u ) d u .

Solution Summary: The author explains that the initial value problem ly

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To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

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Chapter 13 Solutions

To prove: The initial value problem y ″ ( x ) + y ( x ) = 6 x y ( 0 ) = 1 , y ′ ( 0 ) = − 1 is equivalent to the integral equation y ( x ) = x 3 − x + 1 + ∫ 0 x ( u − x ) y ( u ) d u .

Solution Summary: The author explains that the initial value problem ly

Concept explainers

Videos

To prove: The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.

Want to see the full answer?

Chapter 13 Solutions

To prove:

The initial value problem y″(x)+y(x)=6xy(0)=1,y′(0)=−1 is equivalent to the integral equation y(x)=x3−x+1+∫0x(u−x)y(u)du.