I have to solve this ODE:
$$ (y-2x)\frac{dy}{dx}=3y-6x+1 $$
I have to make it in a form of separable variable equation and for that I have tried the following substitutions: $$ u=y-2x $$ and $$ u=3y-6x+1 $$
But none made it "solvable", for example in the first I get to a point where I have this: $$ \frac{e^{y-2x}}{|y-2x|}=x+K $$ where K is any constant. At this point I am unable to solve for y.
Can anyone help me solve this?
On
Starting from Isham's answer.
Using $$3x+c=y-\ln(y-2x+1)$$ the explicit solution is given by $$y=-W\left(-e^{-(x+c+1)}\right)+2 x-1$$ where appears Lambert function.
Using the condition $y(x_0)=y_0$, we get $$c=-\log \left(-(2 x_0-y_0-1)\, e^{(2 x_0-y_0-1)}\right)-x_0-1$$ and then $$y=-W\left((2 x_0-y_0-1) e^{(3 x_0-y_0-x-1)}\right)+2 x-1$$
$$(y-2x)\frac{dy}{dx}=3y-6x+1$$ Substitute $y-2x=z \implies y'-2=z'$ $$z(z'+2)=3z+1$$ $$zz'=z+1$$ $$\int \frac {zdz}{z+1}=x+K$$ $$z-\ln (z+1)=x+K$$ $$y -\ln(y-2x+1)=3x+K$$ Implicit form is correct too. So you can keep it this way
Or consider x as a function of y $$x(y)=\frac 13(y-\ln(y-2x+1))+c$$