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