Explanation Theorem used: 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 . Calculation: The differential equation is, x y ′ + y − y 2 e 2 x = 0 , y ( 1 ) = 2 Rewrite the differential equation as shown below. x y ′ + y − y 2 e 2 x = 0 x d y d x + y − y 2 e 2 x = 0 x d y + ( y − y 2 e 2 x ) d x = 0 Compare the above equation with the equation M ( x , y ) d x + N ( x , y ) d y = 0 and the following is obtained. M ( x , y ) = y − y 2 e 2 x N ( x , y ) = x Differentiate M ( x , y ) with respect to y as shown below. M y ( x , y ) = ∂ ∂ y ( y − y 2 e 2 x ) = 1 − 2 y e 2 x Differentiate N ( x , y ) with respect to x as shown below. N x ( x , y ) = ∂ ∂ x ( x ) = 1 Note that as ∂ M ∂ y ( x , y ) ≠ ∂ N ∂ x ( x , y ) , the given differential equation is not exact. Obtain the integrating factor to make the differential equation exact as shown below. R ( y ) = ∂ N ∂ x − ∂ M ∂ y x M ( x , y ) − y N ( x , y ) = 1 − ( 1 − 2 y e 2 x ) x ( y − y 2 e 2 x ) − y ( x ) = 1 − 1 + 2 y e 2 x x y − x y 2 e 2 x − x y = 2 y e 2 x − x y 2 e 2 x = 2 − x y y e 2 x y e 2 x = − 2 x y Obtain the integrating factor as shown below. μ ( y ) = e ∫ R ( y ) d y = e ∫ − 2 x y d y = e − 2 ln ( x y ) = ( x y ) − 2 = 1 x 2 y 2 Multiply both sides of the equation and the following is obtained. ( x − 2 y − 2 ) ( x d y ) + ( x − 2 y − 2 ) ( y − y 2 e 2 x ) d x = 0 ( y − y 2 e 2 x x 2 y 2 ) d x + ( x x 2 y 2 ) d y = 0 ( y ( 1 − y e 2 x ) x 2 y 2 ) d x + ( x x 2 y 2 ) d y = 0 ( 1 − y e 2 x x 2 y ) d x + ( 1 x y 2 ) d y = 0 Now the functions M ( x , y ) and N ( x , y ) becomes as shown below

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2, Problem 32MP

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The solution of the differential equation xy′+y−y2e2x=0,y(1)=2.

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Chapter 2, Problem 31MP

Chapter 2, Problem 33MP

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Chapter 2 Solutions

Elementary Differential Equations

Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.1 - Prob. 10P

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The solution of the differential equation x y ′ + y − y 2 e 2 x = 0 , y ( 1 ) = 2 .

The solution of the differential equation xy′+y−y2e2x=0,y(1)=2.

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Chapter 2 Solutions