Answer
$$y = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{x}{3}} \right) + 2$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{2}{{9 + {x^2}}} \cr
& {\text{Separate the variables}} \cr
& dy = \frac{2}{{9 + {x^2}}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\frac{2}{{9 + {x^2}}}} dx \cr
& y = 2\left( {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right) + C \cr
& y = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{x}{3}} \right) + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {0,2} \right) \cr
& 2 = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{0}{3}} \right) + C \cr
& C = 2 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{x}{3}} \right) + 2 \cr
& \cr
& {\text{Graph}} \cr} $$
