Answer
$${\text{The answer is option }}\left( {\bf{b}} \right)$$
Work Step by Step
$$\eqalign{
& y' - 2xy = x \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} - 2xy = x \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = - 2x{\text{ }}Q\left( x \right) = x \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{ - \int {2xdx} }} = {e^{ - {x^2}}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{ - {x^2}}}\frac{{dy}}{{dx}} - 2x{e^{ - {x^2}}}y = x{e^{ - {x^2}}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{e^{ - {x^2}}}y} \right] = x{e^{ - {x^2}}} \cr
& d\left[ {{e^{ - {x^2}}}y} \right] = x{e^{ - {x^2}}}dx \cr
& {\text{Integrate both sides}} \cr
& {e^{ - {x^2}}}y = - \frac{1}{2}\int {\left( { - 2x} \right){e^{ - {x^2}}}} dx \cr
& {e^{ - {x^2}}}y = - \frac{1}{2}{e^{ - {x^2}}} + C \cr
& {\text{Solve for }}y \cr
& y = - \frac{1}{2} + C{e^{{x^2}}} \cr
& {\text{The answer is option }}\left( {\bf{b}} \right) \cr} $$
