Answer
$$y = C{e^{2t}} - 3$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = 2y + 6 \cr
& {\text{separate the variables}} \cr
& \frac{{dy}}{{2y + 6}} = dt \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{2dy}}{{2y + 6}}} = 2\int {dt} {\text{ }} \cr
& {\text{ln}}\left| {2y + 6} \right| = 2t + c \cr
& {\text{solve for }}y \cr
& {e^{{\text{ln}}\left| {2y + 6} \right|}} = {e^{2t + c}} \cr
& 2y + 6 = {e^c}{e^{2t}} \cr
& 2y = {e^c}{e^{2t}} - 6 \cr
& y = \frac{{{e^c}{e^{2t}}}}{2} - \frac{6}{2} \cr
& {\text{set }}\frac{{{e^c}}}{2} = C \cr
& y = C{e^{2t}} - 3 \cr} $$
