First Order ODE/(x + y + 4) over (x + y - 6)

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Theorem

The first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x + y - 6}$

has the general solution:

$y - x = 5 \, \map \ln {x + y - 1} + C$


Proof

We note that $(1)$ is in the form:

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

where:

$a e = b d = 1$

Hence we can use First Order ODE in form $y' = \map F {\dfrac{a x + b y + c} {d x + e y + f} }$ where $a e = b d$.

Let:

$z = x + y$

to obtain:

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

where:

$a = 1$
$c = 4$
$d = 1$
$f = -6$

which gives:

\(\ds \dfrac {\d z} {\d x}\) \(=\) \(\ds \dfrac {z + 4} {z - 6} + 1\)
\(\ds \) \(=\) \(\ds \frac {z + 4 + z - 6} {z - 6}\)
\(\ds \) \(=\) \(\ds \frac {2 \paren {z - 1} } {z - 6}\)
\(\ds \leadsto \ \ \) \(\ds \int \rd x\) \(=\) \(\ds \int \frac {z - 6} {2 \paren {z - 1} } \rd z\)


Substitute $v = z - 1$ which gives:

$\dfrac {\d z} {\d v} = 1$

and thence:

$v = x + y - 1$:
\(\ds \int \rd x\) \(=\) \(\ds \frac 1 2 \int \frac {v - 5} v \rd v\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac v 2 - \frac 5 2 \ln v + C_1\)
\(\ds \leadsto \ \ \) \(\ds 2 x\) \(=\) \(\ds v - 5 \ln v + C_2\)


Substituting $x + y - 1$ for $v$:

\(\ds 2 x\) \(=\) \(\ds x + y - 1 - 5 \, \map \ln {x + y - 1} + C_2\)
\(\ds \leadsto \ \ \) \(\ds x - y\) \(=\) \(\ds -5 \, \map \ln {x + y - 1} + C_3\)
\(\ds \leadsto \ \ \) \(\ds y - x\) \(=\) \(\ds 5 \, \map \ln {x + y - 1} + C\)

$\blacksquare$


Sources

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