Answer
$$y = \frac{4}{x}$$
Work Step by Step
$$\eqalign{
& y' + \left( {\frac{1}{x}} \right)y = 0,{\text{ }}y\left( 2 \right) = 2 \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + \left( {\frac{1}{x}} \right)y = 0 \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \frac{1}{x}{\text{ }}Q\left( x \right) = 0 \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\frac{1}{x}dx} }} = {e^{\ln \left| x \right|}} = x \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& x\frac{{dy}}{{dx}} + x\left( {\frac{1}{x}} \right)y = 0 \cr
& x\frac{{dy}}{{dx}} + y = 0 \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {xy} \right] = 0 \cr
& {\text{Integrate both sides}} \cr
& xy = C \cr
& {\text{Solve for }}y \cr
& y = \frac{C}{x}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 2 \right) = 2 \cr
& 2 = \frac{C}{2} \cr
& C = 4 \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{4}{x} \cr} $$
