Explanation Approach: a) An equation is an exact equation when M ( x , y ) d x + N ( x , y ) d y = 0 and ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) holds for all ( x , y ) in R and if the compatibility condition is applicable. b) If M d x + N d y = 0 is exact, the partial derivative ∂ F ∂ y = N . Integrate the equation with respect to y . F ( x , y ) = ∫ N ( x , y ) d y + h ( x ) c) To determine h ( x ) , take the partial derivative with respect to x on both the sides of the equation and substitute ∂ F ∂ x for M . d) Integrate h ' ( x ) to obtain h ( x ) up to a numerical value and substitute in the equation mentioned in point (b) to find the value of F ( x , y ) . e) The solution of the equation M d x + N d y = 0 is given by F ( x , y ) = C . Calculation: Consider the given differential equation. ( y e x y − 1 y ) d x + ( x e x y + x y 2 ) d y = 0 Compare the equation ( y e x y − 1 y ) d x + ( x e x y + x y 2 ) d y = 0 with the standard equation for the exact equation. M d x + N d y = 0 From the above comparison, M = ( y e x y − 1 y ) N = ( x e x y + x y 2 ) Check for the compatibility condition for the equation to be the exact equation. Take the partial differentiation of M with respect to y . ∂ M ∂ y = ∂ ∂ y ( y e x y − 1 y ) = y e x y x + e x y + 1 y 2 = x y e x y + e x y + 1 y 2 Take the partial differentiation of N with respect to x . ∂ N ∂ x = ∂ ∂ x ( x e x y + x y 2 ) = x e x y y + e x y + 1 y 2 = x y e x y + e x y + 1 y 2 From the above, it is clear that the value of ∂ N ∂ x and ∂ M ∂ y are equal. ∂ M ∂ y = ∂ N ∂ x x y e x y + e x y + 1 y 2 = x y e x y + e x y + 1 y 2 Therefore, the equation ( y e x y − 1 y ) d x + ( x e x y + x y 2 ) d y = 0 is exact

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 2.4, Problem 22E

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The solution of the equation (yexy−1y)dx+(xexy+xy2)dy=0 if the initial value is y(1)=1.

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Chapter 2.4, Problem 21E

Chapter 2.4, Problem 23E

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Fundamentals of Differential Equations and Boundary Value Problems

Ch. 2.2 - Prob. 1E Ch. 2.2 - In Problems 1-6, determine whether the given...Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E

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The solution of the equation ( y e x y − 1 y ) d x + ( x e x y + x y 2 ) d y = 0 if the initial value is y ( 1 ) = 1 .

The solution of the equation (yexy−1y)dx+(xexy+xy2)dy=0 if the initial value is y(1)=1.

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