Explanation Approach: Step 1. Calculate the integrating factor. a).Suppose the first partial derivatives of M ( x , y ) and N ( x , y ) are continuous in a rectangle R , then, M ( x , y ) d x + N ( x , y ) d y = 0 Is an exact equation in R if and only if the compatibility condition ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) , holds for all ( x , y ) in R , otherwise M d x + N d y = 0 is not exact. b). If the equation, M ( x , y ) d x + N ( x , y ) d y = 0 Is not exact, but the equation, μ ( x , y ) M ( x , y ) d x + μ ( x , y ) N ( x , y ) d y = 0 That results from multiplying by μ ( x , y ) is exact is called an integrating factor of the equation c).The expression to calculate the integrating factor when it depends only on x is, μ ( x ) = exp [ ∫ ( ∂ M / ∂ y − ∂ N / ∂ x N ) d x ] The expression to calculate the integrating factor when it depends only on y is, μ ( y ) = exp [ ∫ ( ∂ N / ∂ x − ∂ M / ∂ y M ) d y ] Step 2. Method for solving exact equation. a. If M d x + N d y = 0 is exact, then ∂ F ∂ y = N . Integrate this equation with respect to y , F ( x , y ) = ∫ N ( x , y ) d y + h ( x ) b. To determine h ( x ) , take the partial derivative with respect to both sides of the equation and substitute ∂ F ∂ x for M . c. Integrate h ' ( x ) to obtain h ( x ) up to a numerical value and substitute in equation mentioned in point (a) to find the value of F ( x , y ) . d. The solution of M d x + N d y = 0 is given by F ( x , y ) = C . Calculation: Consider the given expression. ( 2 y 2 − 6 x y ) d x + ( 3 x y − 4 x 2 ) d y = 0... ( 1 ) Compare equation ( 1 ) with equation, M d x + N d y = 0 Therefore, M = ( 2 y 2 − 6 x y ) N = ( 3 x y − 4 x 2 ) Partially differentiate M with respect to y . ∂ M ∂ y = ∂ ∂ y ( 2 y 2 − 6 x y ) = 4 y − 6 x Partially differentiate N with respect to x . ∂ N ∂ x = ∂ ∂ x ( 3 x y − 4 x 2 ) = 3 y − 8 x Apply the check for compatibility condition. ∂ M ∂ y = ∂ N ∂ x 4 y − 6 x ≠ 3 y − 8 x Hence, equation ( 1 ) is not an exact differential equation. To make the given equation exact multiply equation ( 1 ) with integrating factor x n y m . x n y m ( 2 y 2 − 6 x y ) d x + x n y m ( 3 x y − 4 x 2 ) d y = 0 ( 2 y m + 2 x n − 6 x n + 1 y m + 1 ) d x + ( 3 x n + 1 y m + 1 − 4 x n + 2 y m ) d y = 0 Now, M = 2 y m + 2 x n − 6 x n + 1 y m + 1 N = ( 3 x n + 1 y m + 1 − 4 x n + 2 y m ) Partially differentiate M with respect to y . ∂ M ∂ y = ∂ ∂ y ( 2 y m + 2 x n − 6 x n + 1 y m + 1 ) = ∂ ∂ y ( 2 y m + 2 x n ) − ∂ ∂ y ( 6 x n + 1 y m + 1 ) = 2 ( m + 2 ) x n y m + 1 − 6 ( m + 1 ) x n + 1 y m Partially differentiate N with respect to x . ∂ N ∂ x = ∂ ∂ x ( 3 x n + 1 y m + 1 − 4 x n + 2 y m ) = ∂ ∂ x ( 3 x n + 1 y m + 1 ) − ∂ ∂ x ( 4 x n + 2 y m ) = 3 ( n + 1 ) x n y m + 1 − 4 ( n + 2 ) x n + 1 y m After the application of the integrating factor the equation is now 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 2.5, Problem 13E

To determine

The integrating factor of the form xnym of the equation (2y2−6xy)dx+(3xy−4x2)dy=0 and solve it.

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Chapter 2.5, Problem 12E

Chapter 2.5, Problem 14E

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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 integrating factor of the form x n y m of the equation ( 2 y 2 − 6 x y ) d x + ( 3 x y − 4 x 2 ) d y = 0 and solve it.

The integrating factor of the form xnym of the equation (2y2−6xy)dx+(3xy−4x2)dy=0 and solve it.

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Chapter 2 Solutions