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

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