First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d/Formulation 1
Jump to navigation
Jump to search
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:
- $z = a x + b y$
to obtain:
- $\dfrac {\d z} {\d x} = b \map F {\dfrac {a z + a c} {d z + f} } + a$
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:
\(\ds a x + b y + c\) | \(=\) | \(\ds a z - a r y + b y + c\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds d x + e y + f\) | \(=\) | \(\ds d z - d r y + \frac {b d} a y + f\) | noting that $e = \dfrac {b d} a$ |
To make $(1)$ separable: we make:
- $b = a r$
and:
- $d r = \dfrac {b d} a$
which comes to the same thing: that $r = \dfrac b a$.
So, we can make the substitution:
- $z = a x + b y$
so:
- $\dfrac {\d z} {\d x} = a + b \dfrac {\d y} {\d x}$
which leaves us with:
\(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds b \map F {\dfrac {z + c} {\frac d a z + f} } + a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b \map F {\dfrac {a z + a c} {d z + f} } + a\) |
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): differential equation: differential equations of the first order and first degree: $(4)$ Equations reducible to homogeneous form
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): differential equation: differential equations of the first order and first degree: $(4)$ Equations reducible to homogeneous form