Explanation Separate the variables and integrate the given differential equation to determine the general solution. Then use the sine addition formula to prove the form of the general solution Given: y = ( cos C ) x + sin C 1 + x 2 Calculation: Given y ' = 1 − y 2 / ( 1 + x 2 ) = > d y 1 − y 2 = d x 1 + x 2 Integrating gives ∫ d y 1 − y 2 = ∫ d x 1 + x 2 = > sin − 1 y = tan − 1 x + C = > y ( x ) = sin ( tan − 1 x + C ) To verify the given form, we use the addition formula sin ( A + B ) = sin A cos B + sin B cos A Therefore, y ( x ) = sin ( tan − 1 x + C

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Differential Equations

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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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Chapter 10, Problem 48CRE

Chapter 10, Problem 50CRE

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Chapter 10 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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Prob. 3PQ Ch. 10.3 - Prob. 4PQ Ch. 10.3 - Prob. 1E Ch. 10.3 - Prob. 2E Ch. 10.3 - Prob. 3E Ch. 10.3 - Prob. 4E Ch. 10.3 - Prob. 5E Ch. 10.3 - Prob. 6E Ch. 10.3 - Prob. 7E Ch. 10.3 - Prob. 8E Ch. 10.3 - Prob. 9E Ch. 10.3 - Prob. 10E Ch. 10.3 - Prob. 11E Ch. 10.3 - Prob. 12E Ch. 10.3 - Prob. 13E Ch. 10.3 - Prob. 14E Ch. 10.3 - Prob. 15E Ch. 10.3 - Prob. 16E Ch. 10.3 - Prob. 17E Ch. 10.3 - Prob. 18E Ch. 10.3 - Prob. 19E Ch. 10.3 - Prob. 20E Ch. 10.3 - Prob. 21E Ch. 10.3 - Prob. 22E Ch. 10.3 - Prob. 23E Ch. 10.3 - Prob. 24E Ch. 10.3 - Prob. 25E Ch. 10.3 - Prob. 26E Ch. 10.3 - Prob. 27E Ch. 10.3 - Prob. 28E Ch. 10.3 - Prob. 29E Ch. 10.4 - Prob. 1PQ Ch. 10.4 - Prob. 2PQ Ch. 10.4 - Prob. 3PQ Ch. 10.4 - Prob. 1E Ch. 10.4 - Prob. 2E Ch. 10.4 - Prob. 3E Ch. 10.4 - Prob. 4E Ch. 10.4 - Prob. 5E Ch. 10.4 - Prob. 6E Ch. 10.4 - Prob. 7E Ch. 10.4 - Prob. 8E Ch. 10.4 - Prob. 9E Ch. 10.4 - Prob. 10E Ch. 10.4 - Prob. 11E Ch. 10.4 - Prob. 12E Ch. 10.4 - Prob. 13E Ch. 10.4 - Prob. 14E Ch. 10.4 - Prob. 15E Ch. 10.4 - 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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 $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$

Want to see the full answer?

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.

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That y=sin(tan−1x+C) is the general solution of y'=1−y2/(1+x2) Then use the addition formula for the sine function to show that general solution may be written as y=(cosC)x+sinC1+x2

Want to see the full answer?

Chapter 10 Solutions

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 $y = \frac{(\cos C) x + \sin C}{\sqrt{1 + x^{2}}}$