Explanation Approach: a) An equation is an exact equation when M ( x , y ) d x + N ( x , y ) d y = 0 and ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) holds for all ( x , y ) in R and if the compatibility condition is applicable. b) An equation is separable when it can be transformed into d y d x = g ( x ) p ( y ) . c) An equation is linear when it can be transformed into a 1 ( x ) d y d x + a 0 ( x ) y = b ( x ) , where a 1 ( x ) , a 0 ( x ) and b ( x ) depends on the independent variable x only. Calculation: Check for the exact equation. Consider the given differential equation. ( x 2 y + x 4 cos x ) d x − x 3 d y = 0 Compare the equation with the standard equation for the exact equation. M d x + N d y = 0 From the above comparison, M = x 2 y + x 4 cos x N = − x 3 Check for the compatibility condition for the equation to be the exact equation. Take the partial differentiation of M with respect to y . ∂ M ∂ y = ∂ ∂ y ( x 2 y + x 4 cos x ) = ∂ ∂ y ( x 2 y ) + ∂ ∂ y ( x 4 cos x ) = x 2 + 0 = x 2 Take the partial differentiation of N with respect to x . ∂ N ∂ x = ∂ ∂ x ( − x 3 ) = − 3 x 2 From the above, it is clear that the value of ∂ N ∂ x and ∂ M ∂ x are not equal. ∂ M ∂ y = ∂ N ∂ x x 2 ≠ − 3 x 2 Therefore, the equation ( x 2 y + x 4 cos x ) d x − x 3 d y = 0 is not an exact equation

7th Edition

ISBN: 9780321977175

Author: Nagle

Publisher: PEARSON CO

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Chapter 2.4, Problem 1E

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To classify:

The equation (x2y+x4cosx)dx−x3dy=0 is separable, linear, exact, or none of these.

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Chapter 2.3, Problem 40E

Chapter 2.4, Problem 2E

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Chapter 2 Solutions

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To classify: The equation ( x 2 y + x 4 cos x ) d x − x 3 d y = 0 is separable, linear, exact, or none of these.

To classify: The equation (x2y+x4cosx)dx−x3dy=0 is separable, linear, exact, or none of these.

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Chapter 2 Solutions

To classify:

The equation (x2y+x4cosx)dx−x3dy=0 is separable, linear, exact, or none of these.