Get the answer to "Integrating factor for xydx+(2x2+3y2−20)dy=0"

Name: Integrating factor for xydx+(2x2+3y2−20)dy=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: Integrating factor for xydx+(2x2+3y2−20)dy=0

Kathy Williams

Answered question

2021-12-20

Integrating factor for

x y d x + (2 x^{2} + 3 y^{2} - 20) d y = 0

Answer & Explanation

Karen Robbins

Beginner2021-12-21Added 49 answers

Given:

$x y d x + (2 x^{2} + 3 y^{2} - 20) d y = 0$
$m d x + n d y = 0$
$\frac{\partial m}{\partial y} = x \frac{\partial n}{\partial x} = 4 x$
Since:

$\frac{\partial m}{\partial y} \neq \frac{\partial n}{\partial x}$
So

$m (y) = \frac{\frac{\partial n}{\partial x} - \frac{\partial m}{\partial y}}{m} = \frac{4 x - x}{x y} = \frac{3}{y}$
And:

$I . f = e^{\int m (y) d y} = e^{\int \frac{3}{4}}$
$I . f = e^{3 \ln y} = e^{{\ln y}^{3}}$
$I . f = y^{3}$

Andrew Reyes

Beginner2021-12-22Added 24 answers

Equation is

$M (x, y) d x + N (x, y) d y = 0 w i t h M = x y, N = 2 x^{2} + 3 y^{2} - 20$ .
it is

$M_{y} = x N_{x} = 4 x, b u t \frac{N_{x} - M_{y}}{M} = \frac{3}{y}$ ,

if only y is a factor, then the integrating factor is

$I . F . = e^{3 \ln y} = y^{3}$

and generates the precise differential.

$d F (x, y) = P d x + Q d y = 0$ , with $P (x, y) = x y^{4}, Q (x, y) = 2 x^{2} y^{3} + 3 y^{6} - 20 y^{3}$ .
Solution

$F (x, y) = C$

by integrating the equation.

$F_{x} = P$

wich gives
$F (x, y) = \frac{x^{2} y^{4}}{2} + G (y)$ ,

Using the

$F_{y} = Q$

$G^{'} (y) = 3 y^{5} - 20 y^{3}$
i.e. $G (y) = (\frac{1}{2}) y^{6} - 5 y^{4} + C$ .

Hence $F (x, y) = (y^{4}) (\frac{x^{2}}{2} + y^{2} - 5) = C$ .

RizerMix

Expert2021-12-29Added 656 answers

The equation is $M (x, y) d x + N (x, y) d y = 0 w i t h M = x y, N = 2 x^{2} + 3 y^{2} - 20.$ it is $M_{y}$ ... The solution $F (x, y) = C$ is obtained integrating the eq.

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