$(x + y)\cfrac{{dy}}{{dx}} = 1$

: The given equation is $\cfrac{{dy}}{{dx}} - x = y$

…(1)
which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Py = Q$

Where, $P = - 1$ and $Q = y$

$\therefore {\rm{I}}.{\rm{F}}. = {e^{\int {Pdy} }} = {e^{\int { - dy} }} = {e^{ - y}}$

$\therefore$ The solution is $x \cdot ({\rm{I}}.{\rm{F}}.) = \int Q \cdot ({\rm{I}}.{\rm{F}}.)dy + C$

$\Rightarrow x \cdot {e^{ - y}} = \int y \cdot {e^{ - y}}dy + C = - y \cdot {e^{ - y}} - \int { - {e^{ - y}}dy} + C$

$= - y{e^{ - y}} + \int {{e^{ - y}}} dy + C = - y{e^{ - y}} + \cfrac{{{e^{ - y}}}}{{ - 1}} + C$

$\Rightarrow x = - y - 1 + C{e^y} \Rightarrow x + y + 1 = C{e^y}$

which is the required solution.