How can I make this 1st order ODE separable? $$ y (2+3xy) \,{\rm d} x = x (2-3xy) \,{\rm d} y $$
I tried using the substitution $y=\frac{v}{x}$, but that didn't get me far. Then I tried using the substitution $y=vx$, but that didn't work either. Any help, please?
 A: Put, $xy=v$ then the equation reduces to $\dfrac{dv}{dx}=\dfrac{v}{x}.\dfrac{4}{2-3v}\implies 2\dfrac{dv}{v}-3v=4\dfrac{dx}{x}$
Integrating we get,
$v^2e^{-3v}=cx^4\implies y^2e^{-3xy}=cx^2$,which is the general solution.Here is the graph,

A: You can treat this as a theorem or a Rule:-
If the equation is $Mdx+Ndy=0$ and $Mx-Ny\neq 0$ . And the equation is of the form $yf(xy)dx+xF(xy)dy=0$.
Then $\frac{1}{Mx-Ny}$ is an integrating factor.
Here $Mx-Ny=4xy$.
So Multiplying by $\frac{1}{4xy}$. we have:-
$$\frac{1}{4x}(2+3xy)dx+\frac{1}{4y}(3xy-2)dy=0$$
$$\frac{1}{2x}dx+3ydx+3xdy-\frac{1}{2y}dy=0$$
$$=\frac{1}{2x}dx+3d(xy)-\frac{1}{2y}dy=0$$
Integrating you have:-
$$\frac{1}{2}\ln(\frac{x}{y})+3xy=C$$ as the solution
A: Doing the same as @Kishalay Sarkar for
$$ x (2-3xy) \, y'-y (2+3xy)  =0$$
$$x y=v \implies  (3 v-2) v'+\frac{4 v}{x}=0\implies \frac 4x=\frac {(2-3v)v'}v$$
Integrating
$$4 \log(x)+c_1=2\log(v)-3 v$$ and the solution is given in terms of Lambert function
$$v=-\frac{2}{3} W\left(c_2\, x^2\right)\implies y=-\frac{2}{3x} W\left(c_2\, x^2\right)$$
