Explanation Procedure used: Consider the first order differential equation: M ( x , y ) d x + N ( x , y ) d y = 0 Find out M x and N y The condition for the exact differential is given by, M y = N x Where, M y is ∂ M ∂ y and N x is ∂ N ∂ x . Calculation: Consider the differential equation in the general form. ( cos x sin x − x y 2 ) d x + y ( 1 − x 2 ) d y = 0 (1) Compare and obtain, M ( x , y ) = ( cos x sin x − x y 2 ) (2) N ( x , y ) = y ( 1 − x 2 ) (3) Partially differentiate the equation (2) with respect to y . M y ( x , y ) = − 2 x y Partially differentiate the equation (3) with respect to x . N x ( x , y ) = − 2 x y Here, M y = N x therefore the equation (1) is exact. Integrate M ( x , y ) with respect to d x considering y as a constant and N ( x , y ) with respect to d y considering x as a constant from equation (1)

First Course in Differential Equations (Instructor's)

11th Edition

ISBN: 9781305965775

Author: ZILL

Publisher: CENGAGE L

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Chapter 2.4, Problem 41E

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To explain: The definition of the explicit solution of the IVP (the blue curve in Figure 2.4.1) at (−1, 1).

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Chapter 2.4, Problem 40E

Chapter 2.4, Problem 42E

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First Course in Differential Equations (Instructor's)

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The definition of the explicit solution of the IVP (the blue curve in Figure 2.4.1) at ( − 1 , 1 ) .

Solution Summary: The author explains the definition of the explicit solution of IVP at (-1,1).

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To explain: The definition of the explicit solution of the IVP (the blue curve in Figure 2.4.1) at (−1, 1).

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Chapter 2 Solutions