Explanation Given information: The given differential equation is 2 y d x + ( x − sin y ) d y = 0 . Concept Used: 1) Let M and N have continuous partial derivatives on an open disk R . The differential equation M ( x , y ) d x + N ( x , y ) d y = 0 is exact if and only if ∂ M ∂ y = ∂ N ∂ x . 2) If M y ( x , y ) − N x ( x , y ) N ( x , y ) = h ( x ) is a function of x alone, then e ∫ h ( x ) d x is an integrating factor. 3) If N x ( x , y ) − M y ( x , y ) M ( x , y ) = k ( y ) is a function of y alone, then e ∫ k ( y ) d y is an integrating factor. 4) The equation M ( x , y ) d x + N ( x , y ) d y = 0 is an exact differential equation when there exists a function f of two variables x and y having continuous partial derivatives such that f x ( x , y ) = M ( x , y ) and f y ( x , y ) = N ( x , y ) The general solution of the equation is f y ( x , y ) = C Calculation: First we will compare differential equation 2 y d x + ( x − sin y ) d y = 0 with the M ( x , y ) d x + N ( x , y ) d y = 0 . We have M = 2 y N = x − sin y ∂ M ∂ y = ∂ ( 2 y ) ∂ y ∂ M ∂ y = 2 ∂ N ∂ x = ∂ ( x − sin y ) ∂ x ∂ N ∂ x = 1 Since ∂ M ∂ y ≠ ∂ N ∂ x , hence this differential equation is not exact. Now, we can calculate N x ( x , y ) − M y ( x , y ) M ( x , y ) k ( y ) = 1 − 2 2 y k ( y ) = − 1 2 y The integrating factor is given as: e ∫ k ( y ) d y = e ∫ ( − 1 2 y ) d y e ∫ k ( y ) d y = e − 1 2 ln y e ∫ k ( y ) d y = 1 y If we multiply the given differential equation by 1 y , we will produce the exact differential equation

11th Edition

ISBN: 8220103600781

Author: Edwards

Publisher: YUZU

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Chapter 16.1, Problem 31E

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The integrating factor that is function of x or y alone and use it to find the general solution of the differential equation 2ydx+(x−siny)dy=0.

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Chapter 16.1, Problem 30E

Chapter 16.1, Problem 32E

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Ch. 16.1 - Exactness What does it mean for the...Ch. 16.1 - Integrating Factor When is it beneficial to use an...Ch. 16.1 - Testing for Exactness In Exercises 3-6, determine...Ch. 16.1 - Testing for Exactness In Exercises 3-6, determine...Ch. 16.1 - Prob. 5E Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E Ch. 16.1 - Solving an Exact Differential Equation In...Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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The integrating factor that is function of x or y alone and use it to find the general solution of the differential equation 2 y d x + ( x − sin y ) d y = 0 .

Solution Summary: The author determines the integrating factor that is function of x or y alone and uses it to find the general solution of the differential equation.

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The integrating factor that is function of x or y alone and use it to find the general solution of the differential equation 2ydx+(x−siny)dy=0.

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Chapter 16 Solutions