Solve the differential equation $\frac{dy}{dx} = \frac{2x-y+2}{2x-y+3}$ 
Solve the differential equation $$\frac{dy}{dx} = \frac{2x-y+2}{2x-y+3}$$

Approaches I have tried:


*

*Using an integrating factor to make it an exact equation. I couldn't find a suitable integrating factor though.

*Shift: put $x = X+a$, $y = Y+b$ and rearrange. This wasn't helpful either as I couldn't get it in a form that I could solve.

*Substitution: put $v = 2x-y$. Then $\frac{dv}{dv} = 2 - \frac{dy}{dx}$. After rearranging and solving using separation of variables, I got $v - ln(v+4) = x + c$, where $c$ is a constant. I didn't know how to find $y(x)$ from here. 

 A: How to find the integrating factor, shown in full details :
$$(2x-y+2)dx-(2x-y+3)dy=0$$
Let $F(x,y)$ be the integrating factor, so that we obtain the exact differential of a function $dU(x,y)=\frac{\partial U}{\partial x}dx+\frac{\partial U}{\partial y}dy$
$$\big((2x-y+2)F(x,y)\big)dx+\big(-(2x-y+3)F(x,y)\big)dy=0$$
$$\frac{\partial U}{\partial x}=(2x-y+2)F(x,y)$$
$$\frac{\partial U}{\partial y}=-(2x-y+3)F(x,y)$$
$$\frac{\partial^2 U}{\partial x \partial y}=\frac{\partial^2 U}{\partial y \partial x}=
\frac{\partial (2x-y+2)F}{\partial y}=\frac{\partial (-2x+y-3)F}{\partial x}$$
$$-F+(2x-y+2)\frac{\partial F}{\partial y}=-2F+(-2x+y-3)\frac{\partial F}{\partial x}$$
Only one solution of this PDE is sufficient for our purpose. So, we search a solution on the form $F(x,y)=g(x)h(y)$
$$(2x-y+2)g h'=-g h+(-2x+y-3)g'h$$
$$(2x-y+2)\frac{h'}{h}=-1+(-2x+y-3)\frac{g'}{g}$$
Separating the terms functions of $x$ from those function of $y$ leads to an obvious result :
$$\frac{h'}{h}=-\frac{g'}{g}=1$$
Hense $h=e^y$ and $g=e^{-x}$ , then $F=gh=e^{y-x}$
An integrating factor is $e^{y-x}$ 
$$dU=(2x-y+2)e^{y-x}dx-(2x-y+3)e^{y-x}dy=0$$
The integration leads to :
$$(2x-y+4)e^{y-x}+C=0$$
This implicite equation is the solution of the ODE. We can express $y(x)$ thanks to a special function :
$$(y-2x-4)e^{(y-2x-4)}=C e^{-x-4}=c e^{-x}$$
$$(y-2x-4)=W\big(c e^{-x} \big)$$
$W(X)$ is the Lambert W function, where $X=c e^{-x}$ 
$$y=2x+4+W\big(c e^{-x} \big)$$
A: Hint:
$\dfrac{dy}{dx} = \dfrac{2x-y+2}{2x-y+3}=1-\dfrac{1}{2x-y+3}$
$2x-y+3=z \implies 2-\dfrac{dy}{dx}=\dfrac{dz}{dx} \implies 2-\dfrac{dz}{dx}=\dfrac{dy}{dx}$
$\therefore 2-\dfrac{dz}{dx} = 1-\dfrac{1}{z} \implies \dfrac{dz}{dx}=1+\dfrac{1}{z}$

I should mention that you can also solve an equation of the form,
$$\dfrac{dy}{dx}=F\left(\dfrac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}\right)$$
For this let $a_1x+b_1y+c_1=U$ and $a_2x+b_2y+c_2=V$
$a_2x+b_2y+c_2=V \implies a_2+b_2\dfrac{dy}{dx}=\dfrac{dV}{dx}
> \implies \dfrac{dy}{dx}=
> \dfrac{1}{b_2}\left(\dfrac{dV}{dx}-a_2\right)$
Similarly, $\dfrac{dy}{dx}=
> \dfrac{1}{b_1}\left(\dfrac{dU}{dx}-a_1\right)$
$\therefore
> \dfrac{dy}{dx}=F\left(\dfrac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}\right)\\
> \implies \dfrac{dU}{dx}=a_1+b_1F\left(\dfrac{U}{V}\right)\\
> \implies\left(\dfrac{dU}{dV}\right)\left(\dfrac{dV}{dx}\right)=a_1+b_1F\left(\dfrac{U}{V}\right)\\
> \implies\dfrac{dU}{dV}=\left(\dfrac{a_1+b_1F\left(\dfrac{U}{V}\right)}{a_2+b_2F\left(\dfrac{U}{V}\right)}\right)$
Which can be simplified further by the substitution $U=VT$.

