Explanation Given information: The given differential equation is ( − 2 y 3 + 1 ) d x + ( 3 x y 2 + x 3 ) 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 3 + 1 ) d x + ( 3 x y 2 + x 3 ) d y = 0 with the M ( x , y ) d x + N ( x , y ) d y = 0 . We have M = − 2 y 3 + 1 N = 3 x y 2 + x 3 ∂ M ∂ y = ∂ ( − 2 y 3 + 1 ) ∂ y ∂ M ∂ y = − 6 y 2 ∂ N ∂ x = ∂ ( 3 x y 2 + x 3 ) ∂ x ∂ N ∂ x = 3 y 2 + 3 x 2 Since ∂ M ∂ y ≠ ∂ N ∂ x , hence this differential equation is not exact. Now, we can calculate M y ( x , y ) − N x ( x , y ) N ( x , y ) h ( x ) = − 6 y 2 − 3 y 2 − 3 x 2 3 x y 2 + x 3 h ( x ) = − 3 ( 3 y 2 + x 2 ) x ( 3 y 2 + x 2 ) h ( x ) = − 3 x The integrating factor is given as: e ∫ h ( x ) d x = e ∫ ( − 3 x ) d x e ∫ h ( x ) d x = e − 3 ln x e ∫ h ( x ) d x = 1 x 3 If we multiply the given differential equation by 1 x 3 , 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 16.1, Problem 32E

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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 (−2y3+1)dx+(3xy2+x3)dy=0.

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Chapter 16.1, Problem 31E

Chapter 16.1, Problem 33E

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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 3 + 1 ) d x + ( 3 x y 2 + x 3 ) 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 (−2y3+1)dx+(3xy2+x3)dy=0.

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Chapter 16 Solutions