Answer
$$y = 7{e^{3t}} + 2$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = 3y - 6 \cr
& \frac{{dy}}{{dt}} = 3y - 6 \cr
& {\text{separate the variables}} \cr
& \frac{{dy}}{{3y - 6}} = dt \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{3dy}}{{3y - 6}}} = 3\int {dt} {\text{ }} \cr
& {\text{ln}}\left| {3y - 6} \right| = 3t + c \cr
& {\text{solve for }}y \cr
& {e^{{\text{ln}}\left| {3y - 6} \right|}} = {e^{3t + c}} \cr
& 3y - 6 = {e^c}{e^{3t}} \cr
& 3y = {e^c}{e^{3t}} + 6 \cr
& y = \frac{{{e^c}{e^{3t}}}}{3} + \frac{6}{3} \cr
& {\text{set }}\frac{{{e^c}}}{3} = C \cr
& y = C{e^{3t}} + 2 \cr
& {\text{the initial condition }}y\left( 0 \right) = 9{\text{ implies that}} \cr
& 9 = C{e^0} + 2 \cr
& 9 = C + 2 \cr
& C = 7 \cr
& {\text{so the solution of the initial value problem is}} \cr
& y = 7{e^{3t}} + 2 \cr} $$
