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

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

such that:
 * $ a e \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:
 * $\dfrac {\mathrm d w} {\mathrm d z} = F \left({\dfrac {a z + b w} {d z + e w} }\right)$

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

Proof
We have:
 * $\dfrac {\mathrm d y} {\mathrm d x} = F \left({\dfrac {a x + b y + c} {d x + e y + f} }\right)$

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

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

Thus:

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

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

Letting:
 * $M \left({z, w}\right) = a z + b w$
 * $N \left({z, w}\right) = d z + e w$

we see:
 * $M \left({tz, tw}\right) = a t z + b t w = t \left({a z + b w}\right) = t M \left({z, w}\right)$
 * $N \left({tz, tw}\right) = d t z + e t w = t \left({d z + e w}\right) = t N \left({z, w}\right)$

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