Answer
$$u = 9{e^{2x - 2}} - 3$$
Work Step by Step
$$\eqalign{
& \frac{{du}}{{dx}} = 2u + 6 \cr
& {\text{separate the variables}} \cr
& \frac{{du}}{{2u + 6}} = dx \cr
& \frac{{2du}}{{2u + 6}} = 2dx \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{2du}}{{2u + 6}}} = 2\int {dx} {\text{ }} \cr
& {\text{ln}}\left| {2u + 6} \right| = 2x + c \cr
& {\text{solve for }}u \cr
& {e^{{\text{ln}}\left| {2u + 6} \right|}} = {e^{2x + c}} \cr
& 2u + 6 = {e^c}{e^{2x}} \cr
& 2u = {e^c}{e^{2x}} - 6 \cr
& u = \frac{{{e^c}{e^{2x}}}}{2} - \frac{6}{2} \cr
& u = \frac{{{e^c}}}{2}{e^{2x}} - 3 \cr
& {\text{set }}\frac{{{e^c}}}{2} = C \cr
& u = C{e^{2x}} - 3 \cr
& {\text{the initial condition }}u\left( 1 \right) = 6{\text{ implies that}} \cr
& 6 = C{e^{2\left( 1 \right)}} - 3 \cr
& 6 = C{e^2} - 3 \cr
& C = \frac{9}{{{e^2}}} \cr
& {\text{so the solution of the initial value problem is}} \cr
& u = \frac{9}{{{e^2}}}{e^{2x}} - 3 \cr
& u = 9{e^{2x - 2}} - 3 \cr} $$
