Answer
$$y = \frac{1}{2}{\ln ^2}x - 2$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{\ln x}}{x},{\text{ }}\left( {1, - 2} \right) \cr
& {\text{Separate the variables}} \cr
& dy = \frac{{\ln x}}{x}dx \cr
& {\text{Integrate}} \cr
& \int {dy} = \int {\frac{{\ln x}}{x}} dx \cr
& y = \frac{1}{2}{\ln ^2}x + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {1, - 2} \right) \cr
& - 2 = \frac{1}{2}{\ln ^2}\left( 1 \right) + C \cr
& C = - 2 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{1}{2}{\ln ^2}x - 2 \cr
& \cr
& {\text{Graph}} \cr} $$
