First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d/Formulation 2

Theorem
The first order ODE:
 * $(1): \quad \dfrac {\mathrm d y} {\mathrm d x} = F \left({\dfrac {a x + b y + c} {d x + e y + f} }\right)$

such that:
 * $a e = b d$

can be solved by substituting:
 * $z = d x + e y$

to obtain:
 * $\dfrac {\mathrm d z} {\mathrm d x} = e F \left({\dfrac {b z + e c} {e z + e f} }\right) + d$

which can be solved by the technique of Separation of Variables.

Proof
When $a e = b d$, it is not possible to make the substitutions:


 * $x := z - h$
 * $y := w - k$

where:
 * $h = \dfrac {c e - b f} {a e - b d}$
 * $k = \dfrac {a f - c d} {a e - b d}$

and so to use the technique of First Order ODE in form $y' = F \left({\dfrac {a x + b y + c} {d x + e y + f} }\right)$.

So, we consider what needs to be done to make $(1)$ separable.

Let us make the substitution:
 * $z = x + r y$

Consider what, if any, value of $r$ would make $(1)$ separable.

We have:

To make $(1)$ separable: we make:
 * $e = d r$

and:
 * $a r = \dfrac {a e} d$

which comes to the same thing: that $r = \dfrac e d$.

So, we can make the substitution:
 * $z = d x + e y$

so:
 * $\dfrac {\mathrm d z} {\mathrm d x} = d + e \dfrac {\mathrm d y} {\mathrm d x}$

which leaves us with: