Answer
$$y = 2{e^{ - 2t}} - 2$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = - 2y - 4 \cr
& \frac{{dy}}{{dt}} = - 2y - 4 \cr
& {\text{separate the variables}} \cr
& \frac{{dy}}{{ - 2y - 4}} = dt \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{ - 2dy}}{{ - 2y - 4}}} = - 2\int {dt} {\text{ }} \cr
& {\text{ln}}\left| { - 2y - 4} \right| = - 2t + c \cr
& {\text{solve for }}y \cr
& {e^{{\text{ln}}\left| { - 2y - 4} \right|}} = {e^{ - 2t + c}} \cr
& - 2y - 4 = {e^c}{e^{ - 2t}} \cr
& - 2y = {e^c}{e^{ - 2t}} + 4 \cr
& y = \frac{{{e^c}{e^{ - 2t}}}}{{ - 2}} + \frac{4}{{ - 2}} \cr
& {\text{set }}\frac{{{e^c}}}{{ - 2}} = C \cr
& y = C{e^{ - 2t}} - 2 \cr
& {\text{the initial condition }}y\left( 0 \right) = 0{\text{ implies that}} \cr
& 0 = C{e^0} - 2 \cr
& C = 2 \cr
& {\text{so the solution of the initial value problem is}} \cr
& y = 2{e^{ - 2t}} - 2 \cr} $$