Explanation Given information: The given differential equation is cos y d x − [ 2 ( x − y ) sin y + cos 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 cos y d x − [ 2 ( x − y ) sin y + cos y ] d y = 0 with the M ( x , y ) d x + N ( x , y ) d y = 0 . We have M = cos y N = − [ 2 ( x − y ) sin y + cos y ] ∂ M ∂ y = ∂ ( cos y ) ∂ y ∂ M ∂ y = − sin y ∂ N ∂ x = ∂ ( − 2 ( x − y ) sin y − cos y ) ∂ x ∂ N ∂ x = − 2 sin y 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 ) = − 2 sin y + sin y cos y k ( y ) = − sin y cos y k ( y ) = − tan y The integrating factor is given as: e ∫ k ( y ) d y = e ∫ ( − tan y ) d y e ∫ k ( y ) d y = e ln cos y e ∫ k ( y ) d y = cos y If we multiply the given differential equation by cos y , we will produce the exact differential equation

11th Edition

ISBN: 9781337275378

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 16, Problem 16RE

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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 cosydx−[2(x−y)siny+cosy]dy=0.

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Chapter 16, Problem 15RE

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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 cos y d x − [ 2 ( x − y ) sin y + cos 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 cosydx−[2(x−y)siny+cosy]dy=0.

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Chapter 16 Solutions