Answer
$$y = \frac{2}{{15}}{\left( {5x + 1} \right)^{3/2}} - \frac{{158}}{{15}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \sqrt {5x + 1} \cr
& {\text{distributing the differentials}} \cr
& dy = \sqrt {5x + 1} dx \cr
& dy = {\left( {5x + 1} \right)^{1/2}}dx \cr
& {\text{integrating }} \cr
& \int {dy} = \int {{{\left( {5x + 1} \right)}^{1/2}}} dx \cr
& {\text{multiply and divide by a constant}} \cr
& y = \frac{1}{5}\int {{{\left( {5x + 1} \right)}^{1/2}}\left( 5 \right)} dx \cr
& {\text{integrate using the power rule}} \cr
& y = \frac{1}{5}\left( {\frac{{{{\left( {5x + 1} \right)}^{3/2}}}}{{3/2}}} \right) + C \cr
& {\text{simplify}} \cr
& y = \frac{2}{{15}}{\left( {5x + 1} \right)^{3/2}} + C\,\,\,\left( {\bf{1}} \right) \cr
& {\text{using the initial value problem }}y\left( 3 \right) = - 2 \cr
& - 2 = \frac{2}{{15}}{\left( {5\left( 3 \right) + 1} \right)^{3/2}} + C \cr
& - 2 = \frac{2}{{15}}{\left( {16} \right)^{3/2}} + C \cr
& - 2 = \frac{{128}}{{15}} + C \cr
& - 2 - \frac{{128}}{{15}} = C \cr
& C = - \frac{{158}}{{15}} \cr
& {\text{substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{2}{{15}}{\left( {5x + 1} \right)^{3/2}} - \frac{{158}}{{15}} \cr} $$
