That y = sin ( tan − 1 x + C ) is the general solution of y ' = 1 − y 2 / ( 1 + x 2 ) Then use the addition formula for the sine function to show that general solution may be written as y = ( cos C ) x + sin C 1 + x 2

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Answer 1

Question

Chapter 10, Problem 49CRE

To determine

That $y = \sin (\tan^{- 1} x + C)$ is the general solution of $y' = \sqrt{1 - y^{2}} / (1 + x^{2})$

Then use the addition formula for the sine function to show that general solution may be written as $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

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Answer 2

Question

Chapter 10, Problem 49CRE

To determine

That $y = \sin (\tan^{- 1} x + C)$ is the general solution of $y' = \sqrt{1 - y^{2}} / (1 + x^{2})$

Then use the addition formula for the sine function to show that general solution may be written as $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 10, Problem 49CRE

To determine

That $y = \sin (\tan^{- 1} x + C)$ is the general solution of $y' = \sqrt{1 - y^{2}} / (1 + x^{2})$

Then use the addition formula for the sine function to show that general solution may be written as $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 10, Problem 49CRE

To determine

That $y = \sin (\tan^{- 1} x + C)$ is the general solution of $y' = \sqrt{1 - y^{2}} / (1 + x^{2})$

Then use the addition formula for the sine function to show that general solution may be written as $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 10, Problem 49CRE

To determine

That $y = \sin (\tan^{- 1} x + C)$ is the general solution of $y' = \sqrt{1 - y^{2}} / (1 + x^{2})$

Then use the addition formula for the sine function to show that general solution may be written as $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

Answer 6

Chapter 10, Problem 49CRE

To determine

That $y = \sin (\tan^{- 1} x + C)$ is the general solution of $y' = \sqrt{1 - y^{2}} / (1 + x^{2})$

Then use the addition formula for the sine function to show that general solution may be written as $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

Answer 7

Chapter 10, Problem 49CRE

To determine

That $y = \sin (\tan^{- 1} x + C)$ is the general solution of $y' = \sqrt{1 - y^{2}} / (1 + x^{2})$

Then use the addition formula for the sine function to show that general solution may be written as $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

Answer 8

Expert Solution & Answer

Answer 9

Expert Solution & Answer

Answer 10

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Answer 11

Ch. 10.1 - Prob. 1PQ Ch. 10.1 - Prob. 2PQ Ch. 10.1 - Prob. 3PQ Ch. 10.1 - Prob. 4PQ 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

That y = sin ( tan − 1 x + C ) is the general solution of y ' = 1 − y 2 / ( 1 + x 2 ) Then use the addition formula for the sine function to show that general solution may be written as y = ( cos C ) x + sin C 1 + x 2

Solution Summary: The author explains how to determine that y = sin (tan 1 x + C) is the general solution of the given differential equation.

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That $y = \sin (\tan^{- 1} x + C)$ is the general solution of $y' = \sqrt{1 - y^{2}} / (1 + x^{2})$

Then use the addition formula for the sine function to show that general solution may be written as $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

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Chapter 10 Solutions

That y = sin ( tan − 1 x + C ) is the general solution of y ' = 1 − y 2 / ( 1 + x 2 ) Then use the addition formula for the sine function to show that general solution may be written as y = ( cos C ) x + sin C 1 + x 2

Solution Summary: The author explains how to determine that y = sin (tan 1 x + C) is the general solution of the given differential equation.

Concept explainers

Videos