Explanation Given: The differential equation is, y ( x + y + 1 ) d x + ( x + 2 y ) d y = 0 Procedure used: Compare the given differential equation with the standard form of exact differential equation M d x + N d y = 0 . To check whether the given differential equation is exact or not, check the condition ∂ M ∂ y = ∂ N ∂ x is satisfied or not. If the given differential equation is not exact, then make it exact by using an appropriate integrating factor. Calculate ∂ M ∂ y − ∂ N ∂ x N . If this term is a function of x only or a constant, then integrating factor is e ∫ f ( x ) d x . Standard solution of the exact differential equation: ∫ M d x + ∫ ( t e r m s o f N f r e e f r o m x ) d y = C Calculation: Compare the given differential equation with the standard form of exact differential equation. Thus, M = y ( x + y + 1 ) and N = x + 2 y . Calculate ∂ M ∂ y as follows. ∂ M ∂ y = ∂ ∂ y y ( x + y + 1 ) = ∂ ∂ y ( x y + y 2 + y ) = x + 2 y + 1 Calculate ∂ N ∂ x as follows. ∂ N ∂ x = ∂ ∂ x ( x + 2 y ) = 1 Since ∂ M ∂ y ≠ ∂ N ∂ x , the given differential equation is not exact. Calculate ∂ M ∂ y − ∂ N ∂ x N , by substituting x + 2 y + 1 for ∂ M ∂ y , 1 for ∂ N ∂ x and x + 2 y for N

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 2.4, Problem 32E

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To solve: The given differential equation by finding an appropriate integrating factor.

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Chapter 2.4, Problem 31E

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