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

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Theorem

The first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$

such that:

$a e = b d$


can be solved by substituting:

$z = d x + e y$


to obtain:

$\dfrac {\d z} {\d x} = e \map F {\dfrac {b z + e c} {e z + e f} } + d$

which can be solved by Solution to Separable Differential Equation.


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 $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$.

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:

\(\ds d x + e y + f\) \(=\) \(\ds d z - d r y + e y + f\)
\(\ds \leadsto \ \ \) \(\ds a x + b y + c\) \(=\) \(\ds a z - a r y + \dfrac {a e} d y + c\) noting that $b = \dfrac {a e} d$


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:

$\dfrac {\d z} {\d x} = d + e \dfrac {\d y} {\d x}$

which leaves us with:

\(\ds \frac {\d z} {\d x}\) \(=\) \(\ds e \map F {\dfrac {\frac b e z + c} {z + f} } + d\)
\(\ds \) \(=\) \(\ds e \map F {\dfrac {b z + e c} {e z + e f} } + d\)

$\blacksquare$