Change of variables to make this differential equation separable: $\frac{dy}{dx}=F\left(\frac{ax+by+c}{dx+ey+f}\right)$ I'm trying to define a change of variables to make the next differential equation separable:  $$\frac{dy}{dx}=F\left(\frac{ax+by+c}{dx+ey+f}\right)$$ assuming that $ae-bd=0$. I tried setting $u=ax+by+c$ and $v=dx+ey+f$ to make the right side of the equation homogeneous-like, but then i got confused trying to compute $\frac{dv}{du}$ (using that $ae-bd=0$). I also tried the substitution $u=\frac{ax+by+c}{dx+ey+f}$, and in this case I used the hypothesis that $ae-bd=0$ to deduce the next expresions for the derivatives of $u$:
$$\frac{du}{dx}=\frac{af-cd}{(dx+ey+f)^2}$$ and
$$\frac{du}{dy}=\frac{bf-ce}{(dx+ey+f)^2}$$
but I don't know how to use this info. Any hints will be appreciated.
$$\frac{dy}{dx}=F\left(\frac{ax+by+c}{dx+ey+f}\right)$$
For $e \ne 0$
$$\frac{dy}{dx}=F\left(\frac{aex+bey+ce}{e(dx+ey+f)}\right)$$
Substitute $ae=bd$
$$\frac{dy}{dx}=F\left(\frac{bdx+bey+ce}{e(dx+ey+f)}\right)$$
$$\frac{dy}{dx}=F\left(\frac{b(dx+ey)+ce}{e(dx+ey+f)}\right)$$
Substitute
 $z=dx+ey+f\implies z'=d+ey'$
$$\frac {(z'-d)}e=F\left(\frac{bz+ce-bf}{ez}\right)$$
This last equation is separable
$$z'=e\underbrace{F\left(\frac{bz+ce-bf}{ez}\right)}_{=h(z)}+d$$
$$\int \frac {dz}{eh(z)+d}=\int dx=x+K$$
You can treat also the case $e =0 \implies bd=0$
$$
u = a x + b y + c \to u' = a + b y'\\
v = d x + e y + f \to v' = d + e y'\\
\frac{du}{dv} = \frac{a+b y'}{d+ e y'}\to y' = \frac{d\frac{du}{dv}-a}{b-e\frac{du}{dv}}
$$

hence

$$
\frac{d\frac{du}{dv}-a}{b-e\frac{du}{dv}} = F\left(\frac uv\right)\to \frac{du}{dv}=\frac{a+bF\left(\frac uv\right)}{d+eF\left(\frac uv\right)}
$$

finally making $u = \lambda v\to u' = \lambda' v + \lambda$

$$
\lambda' v + \lambda = \frac{a+bF\left(\lambda\right)}{d+eF\left(\lambda\right)}
$$

so we arrive to the separable form

$$
\frac{dv}{v} = \frac{d\lambda}{\frac{a+bF\left(\lambda\right)}{d+eF\left(\lambda\right)}-\lambda}
$$