Answer
$$y = \frac{1}{{15}}{\tan ^{ - 1}}\frac{{3t}}{5} + \frac{\pi }{{20}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = \frac{1}{{25 + 9{t^2}}},\,\,\,\,y\left( { - \frac{5}{3}} \right) = \frac{\pi }{{30}} \cr
& \cr
& {\text{Separating variables}} \cr
& dy = \left( {\frac{1}{{25 + 9{t^2}}}} \right)dt \cr
& {\text{Integrating both sides of the equation}} \cr
& \int {dy} = \int {\frac{1}{{25 + 9{t^2}}}dt} \cr
& \int {dy} = \frac{1}{3}\int {\frac{3}{{{5^2} + {{\left( {3t} \right)}^2}}}dt} \cr
& y = \frac{1}{3}\left( {\frac{1}{5}{{\tan }^{ - 1}}\frac{{3t}}{5}} \right) + C \cr
& y = \frac{1}{{15}}{\tan ^{ - 1}}\frac{{3t}}{5} + C \cr
& \cr
& {\text{Using the initial condition }}y\left( { - \frac{5}{3}} \right) = \frac{\pi }{{30}} \cr
& \frac{\pi }{{30}} = \frac{1}{{15}}{\tan ^{ - 1}}\frac{{3\left( { - 5/3} \right)}}{5} + C \cr
& \frac{\pi }{{30}} = \frac{1}{{15}}{\tan ^{ - 1}}\left( { - 1} \right) + C \cr
& \frac{\pi }{{30}} = \frac{1}{{15}}\left( { - \frac{\pi }{4}} \right) + C \cr
& \frac{\pi }{{30}} + \frac{\pi }{{60}} = C \cr
& C = \frac{\pi }{{20}} \cr
& \cr
& Then \cr
& y = \frac{1}{{15}}{\tan ^{ - 1}}\frac{{3t}}{5} + \frac{\pi }{{20}} \cr} $$
