How can I make this 1st order ODE separable?

Question

$$ y (2+3xy) \,{\rm d} x = x (2-3xy) \,{\rm d} y $$

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?

Answer 1

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,

Answer 2

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

Answer 3

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 $$v=-\frac{2}{3} W\left(c_2\, x^2\right)\implies y=-\frac{2}{3x} W\left(c_2\, x^2\right)$$