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Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
BuyFindarrow_forward

Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 16.1, Problem 28E
Textbook Problem
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Finding an Integrating Factor In Exercises 23-32, find the integrating factor that is a function of x or y alone and use it to find the general solution of the differential equation.

( 2 x 2 y 1 ) d x + x 3 d y = 0

To determine

The integrating factor that is function of x or y alone and use it to find the general solution of the differential equation (2x2y1)dx+x3dy=0.

Explanation of Solution

Given information:

The given differential equation is (2x2y1)dx+x3dy=0.

Concept Used:

1) Let M and N have continuous partial derivatives on an open disk R. The differential equation

M(x,y)dx+N(x,y)dy=0 is exact if and only if My=Nx.

2) If My(x,y)Nx(x,y)N(x,y)=h(x) is a function of x alone, then eh(x)dx is an integrating factor.

3) If Nx(x,y)My(x,y)M(x,y)=k(y) is a function of y alone, then ek(y)dy is an integrating factor.

4) The equation M(x,y)dx+N(x,y)dy=0 is an exact differential equation when there exists a function f of two variables x and y having continuous partial derivatives such that

fx(x,y)=M(x,y) and fy(x,y)=N(x,y)

The general solution of the equation is fy(x,y)=C

Calculation:

First we will compare differential equation (2x2y1)dx+x3dy=0 with the M(x,y)dx+N(x,y)dy=0.

We have

M=2x2y1N=x3My=(2x2y1)yMy=2x2Nx=(x3)xNx=3x2

Since MyNx, hence this differential equation is not exact

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Chapter 16 Solutions

Multivariable Calculus
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