Answer
$$y = 6{e^{ - 3t}}$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) + 3y = 0,\,\,\,\,\,y\left( 0 \right) = 6 \cr
& {\text{Rewrite}} \cr
& \frac{{dy}}{{dt}} + 3y = 0 \cr
& \frac{{dy}}{{dt}} = - 3y \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{y} = - 3dt \cr
& {\text{Integrating}} \cr
& \int {\frac{{dy}}{y}} = - 3\int {dt} \cr
& \ln \left| y \right| = - 3t + C \cr
& {\text{Use the initial condition }}\,y\left( 0 \right) = 6 \cr
& \ln \left| 6 \right| = - 3t + C \cr
& 6 = - 3\left( 0 \right) + C \cr
& C = \ln 6 \cr
& ,{\text{ then}} \cr
& \ln \left| y \right| = - 3t + \ln 6 \cr
& {\text{Solve for }}y \cr
& {e^{\ln \left| y \right|}} = {e^{ - 3t + \ln 6}} \cr
& y = {e^{\ln 6}}{e^{ - 3t}} \cr
& y = 6{e^{ - 3t}} \cr} $$
