Answer
$$y = \frac{2}{{{x^2}}},{\text{ where }}x > 0$$
Work Step by Step
$$\eqalign{
& {x^2}y' + 2xy = 0,{\text{ }}y\left( 1 \right) = 2 \cr
& {\text{The left - hand side can be interpreted as the derivative of the}} \cr
& {\text{product }}{x^2}y.{\text{ }}\frac{d}{{dx}}\left[ {{x^2}y} \right] = {x^2}y' + 2xy,{\text{ then,}} \cr
& {x^2}y' + 2xy = 0 \cr
& \frac{d}{{dx}}\left[ {{x^2}y} \right] = 0 \cr
& {\text{Integrate both sides with respect to }}x \cr
& {x^2}y = C \cr
& {\text{Solve for }}y \cr
& y = \frac{C}{{{x^2}}} \cr
& {\text{Use the initial condition }}y\left( 1 \right) = 2 \cr
& 2 = \frac{C}{{{{\left( 1 \right)}^2}}} \Rightarrow C = 2 \cr
& {\text{,then}} \cr
& y = \frac{2}{{{x^2}}},{\text{ where }}x > 0 \cr} $$
