Answer
$${\text{The answer is option }}\left( {\bf{a}} \right)$$
Work Step by Step
$$\eqalign{
& y' - 2xy = 0 \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} - 2xy = 0 \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{dx}} = 2xy \cr
& \frac{{dy}}{y} = 2xdx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{{dy}}{y}} = \int {2x} dx \cr
& \ln \left| y \right| = {x^2} + {C_1} \cr
& {e^{\ln \left| y \right|}} = {e^{{x^2}}}{e^{{C_1}}} \cr
& y = C{e^{{x^2}}} \cr
& {\text{The answer is option }}\left( {\bf{a}} \right) \cr} $$
