Explanation Given information: The given differential equation is ( 3 x 2 y 2 ) d x + ( 2 x 3 y − 3 y 2 ) d y = 0 with initial condition y ( 1 ) = 2 . Concept Used: 1) Let M and N have continuous partial derivatives on an open disk R . The differential equation M ( x , y ) d x + N ( x , y ) d y = 0 is exact if and only if ∂ M ∂ y = ∂ N ∂ x 2) The equation M ( x , y ) d x + N ( x , y ) d y = 0 is an exact differential equation when there exists a function f of two variables x and y having continuous partial derivatives such that f x ( x , y ) = M ( x , y ) and f y ( x , y ) = N ( x , y ) The general solution of the equation is f y ( x , y ) = C Calculation: First we will compare differential equation ( 3 x 2 y 2 ) d x + ( 2 x 3 y − 3 y 2 ) d y = 0 with the M ( x , y ) d x + N ( x , y ) d y = 0 . We have M = 3 x 2 y 2 N = 2 x 3 y − 3 y 2 ∂ M ∂ y = ∂ ( 3 x 2 y 2 ) ∂ y ∂ M ∂ y = 6 x 2 y ∂ N ∂ x = ∂ ( 2 x 3 y − 3 y 2 ) ∂ x ∂ N ∂ x = 6 x 2 y Since ∂ M ∂ y = ∂ N ∂ x , hence this differential equation is exact

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ISBN: 9781337604796

Author: Larson

Publisher: CENGAGE L

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Chapter 16, Problem 10RE

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The particular solution of the differential equation (3x2y2)dx+(2x3y−3y2)dy=0 that satisfies the initial condition y(1)=2.

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Chapter 16, Problem 9RE

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Ch. 16.1 - Exactness What does it mean for the...Ch. 16.1 - Integrating Factor When is it beneficial to use an...Ch. 16.1 - Testing for Exactness In Exercises 3-6, determine...Ch. 16.1 - Testing for Exactness In Exercises 3-6, determine...Ch. 16.1 - Prob. 5E Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E Ch. 16.1 - Solving an Exact Differential Equation In...Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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The particular solution of the differential equation ( 3 x 2 y 2 ) d x + ( 2 x 3 y − 3 y 2 ) d y = 0 that satisfies the initial condition y ( 1 ) = 2 .

Solution Summary: The author explains how the differential equation satisfies the initial condition y(1)=2.

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The particular solution of the differential equation (3x2y2)dx+(2x3y−3y2)dy=0 that satisfies the initial condition y(1)=2.

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