Integrating factor for xydx+(2x2+3y2−20)dy=0

Given:

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

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

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

 ${\displaystyle I.f=e^{\int m\left(y\right) dy }=e^{\int \frac{3}{4}}}$ 
${\displaystyle I.f=e^{3\ln y}=e^{{\ln y}^3}}$ 
${\displaystyle I.f=y^3}$

Equation is

 ${\displaystyle M(x,y) dx +N(x,y) dy =0withM=xy,N=2x^2+3y^2-20}$. 
it is

 ${\displaystyle M_y=x{N}_x=4x,but\frac{N_x-M_y}{M}=\frac{3}{y}}$, 

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

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

and generates the precise differential. 

${\displaystyle dF(x,y)=P dx +Q dy =0}$, with ${\displaystyle P(x,y)=x{y}^4,Q(x,y)=2x^2{y}^3+3y^6-20y^3}$. 
Solution 

${\displaystyle F(x,y)=C}$ 

by integrating the equation.

${\displaystyle F_x=P}$ 

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

Using the

 ${\displaystyle F_y=Q}$

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

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

The equation is $M(x,y)dx+N(x,y)dy=0withM=xy,N=2x^2+3y^2-20.$ it is $M_y$... The solution $F(x,y)=C$ is obtained integrating the eq.