First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))
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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:
\(\ds \frac {\d w} {\d z}\) | \(=\) | \(\ds \map F {\frac {a \paren {z - h} + b \paren {w - k} + c} {d \paren {z - h} + e \paren {w - k} + f} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map F {\frac {a z + b w - a h - b k + c} {d z + e w - d h - e k + f} }\) |
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:
\(\ds a h + b k\) | \(=\) | \(\ds c\) | ||||||||||||
\(\ds d h + e k\) | \(=\) | \(\ds f\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a d h + b d k\) | \(=\) | \(\ds c d\) | |||||||||||
\(\ds a d h + a e k\) | \(=\) | \(\ds a f\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds k \paren {a e - b d}\) | \(=\) | \(\ds a f - c d\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds k\) | \(=\) | \(\ds \frac {a f - c d} {a e - b d}\) |
Similarly:
\(\ds a h + b k\) | \(=\) | \(\ds c\) | ||||||||||||
\(\ds d h + e k\) | \(=\) | \(\ds f\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a e h + b e k\) | \(=\) | \(\ds c e\) | |||||||||||
\(\ds b d h + b e k\) | \(=\) | \(\ds b f\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds h \paren {a e - b d}\) | \(=\) | \(\ds c e - b f\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds h\) | \(=\) | \(\ds \frac {c e - b f} {a e - b d}\) |
We note that the above works if and only if:
- $a e - b d \ne 0 \implies a e \ne b d$
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.
$\blacksquare$
Examples
$\dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x - y - 6}$
The first order ODE:
- $(1): \quad \dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x - y - 6}$
has the general solution:
- $\map \arctan {\dfrac {y + 5} {x - 1} } = \ln \sqrt {\paren {x - 1}^2 + \paren {y + 5}^2} + C$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.7$: Homogeneous Equations: Problem $4 \text{(a)}$
- 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