Answer
$$y = 2{e^{ - {x^2} + x}}{\text{ }}$$
Work Step by Step
$$\eqalign{
& y' + \left( {2x - 1} \right)y = 0,{\text{ }}y\left( 1 \right) = 2 \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + \left( {2x - 1} \right)y = 0 \cr
& {\text{Separating the variables}} \cr
& \frac{{dy}}{{dx}} = - \left( {2x - 1} \right)y \cr
& \frac{{dy}}{y} = - \left( {2x - 1} \right)dx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{1}{y}} dy = - \int {\left( {2x - 1} \right)} dx \cr
& \ln \left| y \right| = - \left( {{x^2} - x} \right) + {C_1} \cr
& \ln \left| y \right| = - {x^2} + x + {C_1} \cr
& {\text{Solve for }}y \cr
& {e^{\ln \left| y \right|}} = {e^{ - {x^2} + x}}{e^{{C_1}}} \cr
& y = C{e^{ - {x^2} + x}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 1 \right) = 2 \cr
& 2 = C{e^{ - {{\left( 1 \right)}^2} + \left( 1 \right)}} \cr
& C = 2 \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = 2{e^{ - {x^2} + x}}{\text{ }} \cr} $$
