Answer
$$y = \frac{C}{{2 - x}} + 1$$
Work Step by Step
$$\eqalign{
& \left( {x - 2} \right)y' + y = 1 \cr
& {\text{Subtract }}y{\text{ from both sides of the equation}} \cr
& \left( {x - 2} \right)y' = 1 - y \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \left( {x - 2} \right)\frac{{dy}}{{dx}} = 1 - y \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{1 - y}} = \frac{{dx}}{{x - 2}} \cr
& \frac{{dy}}{{y - 1}} = \frac{{dx}}{{2 - x}} \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{{dy}}{{y - 1}}} = \int {\frac{{dx}}{{2 - x}}} \cr
& \ln \left| {y - 1} \right| = - \ln \left| {2 - x} \right| + {C_1} \cr
& {\text{Solve for }}y \cr
& {e^{\ln \left| {y - 1} \right|}} = {e^{ - \ln \left| {2 - x} \right|}}{e^{{C_1}}} \cr
& y - 1 = \left( {\frac{1}{{2 - x}}} \right)C \cr
& y = \frac{C}{{2 - x}} + 1 \cr
& y = \frac{C}{{2 - x}} + 1 \cr} $$
