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

.: $\cfrac{{dy}}{{dx}} + y = 1 \Rightarrow \cfrac{{dy}}{{dx}} = - (y - 1) \Rightarrow \cfrac{{dy}}{{y - 1}} = - dx$

…(1)
Integrating (1) both sides,

we get
$\int {\cfrac{{dy}}{{y - 1}}} = - \int {dx} \Rightarrow \log (y - 1) = - x + {C_1}$

$\Rightarrow y - 1 = {e^{ - x + {C_1}}} \Rightarrow y = 1 + {e^{ - x}} \cdot {e^{{C_1}}} \Rightarrow y = 1 + C{e^{ - x}}$

Hence, $y = 1 + C{e^{ - x}}$

which is the required solution.

$[C = {e^{{C_1}}}]$