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

Theorem
The first order ODE:
 * $(1): \quad \dfrac {\mathrm d y} {\mathrm d x} = \dfrac {x + y + 4} {x - y - 6}$

has the solution:
 * $\arctan \left({\dfrac{y + 5} {x - 1} }\right) = \ln \sqrt{\left({x - 1}\right)^2 + \left({y + 5}\right)^2} + C$

Proof
We note that $(1)$ is in the form:
 * $\dfrac {\mathrm d y} {\mathrm d x} = F \left({\dfrac {a x + b y + c} {d x + e y + f} }\right)$

where:
 * $a e = -1 \ne b d = 1$

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

Let:
 * $x = z - h$ where $h = 2 / -2 = -1$
 * $y = w - k$ where $k = -10 / -2 = 5$.

Then:


 * $\dfrac {\mathrm d w} {\mathrm d z} = \dfrac {z + w} {z - w}$

From the solution to $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {x + y} {x - y}$:


 * $\arctan \dfrac w z = \ln \sqrt{z^2 + w^2} + C$

We have:
 * $x = z - \left({-1}\right)$
 * $y = w - 5$

and so:
 * $z = x - 1$
 * $w = y + 5$

This gives:
 * $\arctan \left({\dfrac{y + 5} {x - 1} }\right) = \ln \sqrt{\left({x - 1}\right)^2 + \left({y + 5}\right)^2} + C$