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

Theorem
The first order ODE:
 * $\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$

such that:
 * $ a e \ne b d$

can be solved by substituting:
 * $x := z - h$
 * $y := w - k$

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

to obtain:
 * $\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$

which can be solved by the technique of Solution to Homogeneous Differential Equation.

Proof
We have:
 * $\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$

Make the substitutions:
 * $x := z - h$
 * $y := w - k$

We have:
 * $\dfrac {\d x} {\d z} = 1$
 * $\dfrac {\d y} {\d w} = 1$

Thus:

In order to simplify this appropriately, we wish to reduce it to the form:
 * $\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$

by finding values of $h$ and $k$ such that:
 * $a h + b k = c$
 * $d h + e k = f$

So:

Similarly:

We note that the above works :
 * $a e - b d \ne 0 \implies a e \ne b d$

Thus:
 * $(1): \quad \dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$

Letting:
 * $\map M {z, w} = a z + b w$
 * $\map N {z, w} = d z + e w$

we see:
 * $\map M {t z, t w} = a t z + b t w = t \paren {a z + b w} = t \, \map M {z, w}$
 * $\map N {t z, t w} = d t z + e t w = t \paren {d z + e w} = t \, \map N {z, w}$

Thus, by definition, $(1)$ is homogeneous.