Answer
$$y = 2 + C{e^{ - 3x}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} + 3y = 6 \cr
& {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{comparing }}\frac{{dy}}{{dx}} + 3y = 6{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is 3 }} \cr
& {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int 3 dx}} = {e^{3x}} \cr
& {\text{multiplying both sides of the differential equation by }}{e^x} \cr
& {e^{3x}}\frac{{dy}}{{dx}} + 3{e^{3x}}y = 6{e^{3x}} \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {{e^{3x}}y} \right] = 6{e^{3x}} \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& {e^{3x}}y = \int {6{e^{3x}}} dx \cr
& {e^{3x}}y = 6\left( {\frac{{{e^{3x}}}}{3}} \right) + C \cr
& y = \frac{{2{e^{3x}}}}{{{e^{3x}}}} + \frac{C}{{{e^{3x}}}} \cr
& y = 2 + \frac{C}{{{e^{3x}}}} \cr
& {\text{or}} \cr
& y = 2 + C{e^{ - 3x}} \cr} $$
