How can I get integrating factor for $y(2x-y-1)dx+x(2y-x-1)dy=0$

Question

How can I solve this DE using an integrating factor to make it exact DE ? What is the method used to determine the integrating factor , or we just try ?

$y(2x-y-1)dx+x(2y-x-1)dy=0$

Answer 1

$$y(2x-y-1)dx+x(2y-x-1)dy=0 \tag 1$$ I will try another method than integrating factor.

The symmetry draw us to change of variables : $$\begin{cases} s=x+y\\ p=xy \end{cases} \quad\to\quad \begin{cases} x=\frac12\left(s+\sqrt{s^2-4p} \right)\\ y=\frac12\left(s-\sqrt{s^2-4p} \right) \end{cases}$$

$$\begin{cases} dx=\left(\frac12+\frac{s}{2\sqrt{s^2-4p}}\right)ds - \frac{s}{\sqrt{s^2-4p}}dp\\ dy=\left(\frac12-\frac{s}{2\sqrt{s^2-4p}}\right)ds + \frac{s}{\sqrt{s^2-4p}}dp\end{cases}$$

We put the above equations $x,y,dx,dy$ into Eq.$(1)$. This leads to a big equation. After simplification the remaining terms are :

$$3p\,ds-(s+1)dp=0$$ This surprisingly simple ODE is easy to solve : $$(s+1)^3=c\,p$$ The solution of the ODE $(1)$ , expressed on the form of implicit equation, is : $$(x+y+1)^3=c\,xy \tag 2$$ where $c$ is an arbitrary constant.

If you want the explicit form $y(x)$, you will have to solve for $y$ the cubic equation : $$y^3+3(x+1)y^2+(3(x+1)^2-cx)y+(x+1)^3=0$$

NOTE : Since the solution is now known on the very simple form $(2)$, it is probably possible to work backward in order to find an other method of solving which might appears smarter afterwards.

INTEGRATING FACTOR :

An integrating factor for Eq.$(1)$ is :$\quad\frac{(x+y+1)^2}{x^2y^2}$

I must confess that this integrating factor was found in working backward from Eq.$(2)$. Of course, that is a back-door route, not recommended for a straightforward search of integrating factor.