Kathy Williams

2021-12-20

Integrating factor for $xydx+\left(2{x}^{2}+3{y}^{2}-20\right)dy=0$

Karen Robbins

Expert

Given:

Since:

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

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

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

Andrew Reyes

Expert

Equation is

it is

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

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

and generates the precise differential.

, with $P\left(x,y\right)=x{y}^{4},Q\left(x,y\right)=2{x}^{2}{y}^{3}+3{y}^{6}-20{y}^{3}$
Solution

$F\left(x,y\right)=C$

by integrating the equation.

${F}_{x}=P$

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

Using the

${F}_{y}=Q$

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

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

RizerMix

Expert

The equation is it is ${M}_{y}$... The solution $F\left(x,y\right)=C$ is obtained integrating the eq.