The range of h for which the initial value problem, y ′ ( x ) = y 2 ( x ) − x , y ( 1 ) = 0 has unique solution on [ 1 − h , 1 + h ] if, R 1 = { ( x , y ) : | x − 1 | ≤ 1 , | y | ≤ 1 } .

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Answer 1

Question

Chapter 13.2, Problem 9E

To determine

To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.

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Q.1) Classify the following statements as a true or false statements: Q a. A simple ring R is simple as a right R-module. b. Every ideal of ZZ is small ideal. very den to is lovaginz c. A nontrivial direct summand of a module cannot be large or small submodule. d. The sum of a finite family of small submodules of a module M is small in M. e. The direct product of a finite family of projective modules is projective f. The sum of a finite family of large submodules of a module M is large in M. g. Zz contains no minimal submodules. h. Qz has no minimal and no maximal submodules. i. Every divisible Z-module is injective. j. Every projective module is a free module. a homomorp cements Q.4) Give an example and explain your claim in each case: a) A module M which has a largest proper submodule, is directly indecomposable. b) A free subset of a module. c) A finite free module. d) A module contains no a direct summand. e) A short split exact sequence of modules.

Answer 2

Question

Chapter 13.2, Problem 9E

To determine

To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.

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Check out a sample textbook solution

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Answer 3

Question

Chapter 13.2, Problem 9E

To determine

To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.

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Answer 4

Question

Chapter 13.2, Problem 9E

To determine

To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.

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Answer 5

Chapter 13.2, Problem 9E

To determine

To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.

Answer 6

Chapter 13.2, Problem 9E

To determine

To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.

Answer 7

Chapter 13.2, Problem 9E

To determine

To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.

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

The range of h for which the initial value problem, y ′ ( x ) = y 2 ( x ) − x , y ( 1 ) = 0 has unique solution on [ 1 − h , 1 + h ] if, R 1 = { ( x , y ) : | x − 1 | ≤ 1 , | y | ≤ 1 } .

Solution Summary: The author calculates the value of h for which the initial value problem has a unique solution.

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To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.

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Chapter 13 Solutions

The range of h for which the initial value problem, y ′ ( x ) = y 2 ( x ) − x , y ( 1 ) = 0 has unique solution on [ 1 − h , 1 + h ] if, R 1 = { ( x , y ) : | x − 1 | ≤ 1 , | y | ≤ 1 } .

Solution Summary: The author calculates the value of h for which the initial value problem has a unique solution.

Concept explainers

Videos

To find: The range of h for which the initial value problem, y′(x)=y2(x)−x, y(1)=0 has unique solution on [1−h,1+h] if, R1={(x,y):|x−1|≤1,|y|≤1}.

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Chapter 13 Solutions

To find:

The range of h for which the initial value problem,

y′(x)=y2(x)−x, y(1)=0

has unique solution on [1−h,1+h] if,

R1={(x,y):|x−1|≤1,|y|≤1}.