Program Explanation Given information: The differential equation is d y d x = − 3 x 2 + 2 y 2 4 x y . Explanation: The given differential equation can be solved by the method of exact differential equation and by homogeneous form. The given differential equation can be written as, 4 x y d y = − ( 3 x 2 + 2 y 2 ) d x ( 3 x 2 + 2 y 2 ) d x + 4 x y d y = 0 The general form of exact differential equation is, M ( x , y ) d x + N ( x , y ) d y = 0 The above differential equation is exact if and only if the below condition satisfies, ∂ M d y = ∂ N ∂ x ....... (1) Compare the given differential equation with the general differential equation to obtain the value of M ( x , y ) and N ( x , y ) as, M ( x , y ) = 3 x 2 + 2 y 2 N ( x , y ) = 4 x y Substitute 3 x 2 + 2 y 2 for M and 4 x y for N in equation (1) as, ∂ ∂ y ( 3 x 2 + 2 y 2 ) = ? ∂ ∂ x ( 4 x y ) 0 + 4 y = ? 4 y ⋅ 1 4 y = 4 y Thus, the given differential equation is exact. The value of M ( x , y ) can be written as, M ( x , y ) = ∂ F ∂ x The value of F ( x , y ) can be obtained if both the sides of above equation are integrated with respect to x as, ∫ M ( x , y ) d x = F ( x , y ) Substitute 3 x 2 + 2 y 2 in for M ( x , y ) in above equation as, F ( x , y ) = ∫ ( 3 x 2 + 2 y 2 ) d x = 3 ⋅ x 3 3 + 2 y 2 ⋅ x + g ( y ) = x 3 + 2 x y 2 + g ( y ) Now, differentiate the above equation with respect to y , to obtain the value of N ( x , y ) as, N ( x , y ) = ∂ F ∂ y = ∂ ∂ y ( x 3 + 2 x y 2 + g ( y ) ) = 0 + 2 x ⋅ 2 y + g ′ ( y ) = 4 x y + g ′ ( y ) Substitute 4 x y for N ( x , y ) in above equation as, 4 x y = 4 x y + g ′ ( y ) g ′ ( y ) = 0 Integrate the above equation to obtain the value of g ( y ) as, ∫ g ′ ( y ) d y = ∫ 0 d y g ( y ) = 0 + c = c Substitute c for g ( y ) in the value of F ( x , y ) as, F ( x , y ) = x 3 + 2 x y 2 + c Therefore, the solution of the given differential equation is, x 3 + 2 x y 2 = C Here, C is the constant of integration and absorbed the constant c

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

5th Edition

ISBN: 9780134872971

Author: Edwards, C., Penney, David, Calvis

Publisher: PEARSON

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Chapter 1, Problem 33RP

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Program Description: Purpose of problem is to derive general solutions of the differential equation dydx=−3x2+2y24xy in two different ways.

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Chapter 1, Problem 32RP

Chapter 1, Problem 34RP

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Chapter 1 Solutions

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

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Program Description: Purpose of problem is to derive general solutions of the differential equation d y d x = − 3 x 2 + 2 y 2 4 x y in two different ways.

Solution Summary: The author explains that the given differential equation can be solved by the method of exact and by homogeneous form.

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Program Description: Purpose of problem is to derive general solutions of the differential equation dydx=−3x2+2y24xy in two different ways.

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Chapter 1 Solutions