Answer
$$y = 10{e^{2t}} - 2$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = 2y + 4,\,\,\,\,\,y\left( 0 \right) = 8 \cr
& {\text{Rewrite}} \cr
& \frac{{dy}}{{dt}} = 2y + 4 \cr
& \frac{{dy}}{{dt}} = 2\left( {y + 2} \right) \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{y + 2}} = 2dt \cr
& {\text{Integrating}} \cr
& \int {\frac{{dy}}{{y + 2}}} = 2\int {dt} \cr
& \ln \left| {y + 2} \right| = 2t + C \cr
& {\text{Use the initial condition }}\,y\left( 0 \right) = 8 \cr
& \ln \left| {8 + 2} \right| = 2\left( 0 \right) + C \cr
& C = \ln10 \cr
& ,{\text{ then}} \cr
& \ln \left| {y + 2} \right| = 2t + \ln 10 \cr
& {\text{Solve for }}y \cr
& {e^{\ln \left| {y + 2} \right|}} = {e^{2t + \ln 10}} \cr
& y + 2 = {e^{\ln 10}}{e^{2t}} \cr
& y = 10{e^{2t}} - 2 \cr} $$
