Solve $x\ y\ dx=(y^3+x^2y+x^2)\ dy$ differential equation.

Solve $x\ y\ dx=(y^3+x^2y+x^2)\ dy$ differential equation.

What I tried so far.

$M(x,y)=x\ y$ $N(x,y)=y^3+x^2y+x^2$

$\frac{\partial M}{\partial y}=x$ $\frac{\partial N}{\partial x}=2x+2xy$

They are not equal so we need integrating factor.

$$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} }{N}=\frac{2xy-x}{y^3+x^2y+x^2}$$

so $$\mu(x)=e^{-{\huge \int}{\frac{(2xy-x)dx}{y^3+x^2+x^2y}}}$$

from here I can't continue.

EDIT.

I am getting $$\mu=|(1+y)\ x^2+y^3|^\frac{1-2y}{2+2y}$$

Solution 1:

$$xydx=(y^3+x^2y+x^2)dy$$ $$x(ydx-xdy)=y(y^2+x^2)dy$$ $$(ydx-xdy)=\dfrac yx(y^2+x^2)dy$$ Divide by $y^2$: $$d \left (\dfrac {x}y \right)=\dfrac y x \left(1+\dfrac {x^2}{y^2} \right)dy$$ This is separable: $$\dfrac {du}{1+u}=2dy$$ With $u=\left (\dfrac xy\right)^2$.