Answer
$$y = C{e^{3t}} + \frac{4}{3}$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = 3y - 4 \cr
& \frac{{dy}}{{dt}} = 3y - 4 \cr
& {\text{separate the variables}} \cr
& \frac{{dy}}{{3y - 4}} = dt \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{3dy}}{{3y - 4}}} = 3\int {dt} {\text{ }} \cr
& {\text{ln}}\left| {3y - 4} \right| = 3t + c \cr
& {\text{solve for }}y \cr
& {e^{{\text{ln}}\left| {3y - 4} \right|}} = {e^{3t + c}} \cr
& 3y - 4 = {e^c}{e^{3t}} \cr
& 3y = {e^c}{e^{3t}} + 4 \cr
& y = \frac{{{e^c}{e^{3t}}}}{3} + \frac{4}{3} \cr
& {\text{set }}\frac{{{e^c}}}{3} = C \cr
& y = C{e^{3t}} + \frac{4}{3} \cr} $$
