Change of variables to make this differential equation separable: $\frac{dy}{dx}=F\left(\frac{ax+by+c}{dx+ey+f}\right)$

Question

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.

Answer 1

$$\frac{dy}{dx}=F\left(\frac{ax+by+c}{dx+ey+f}\right)$$ 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 $e =0 \implies bd=0$