Answer
$${e^{3x}}y = \frac{1}{6}{e^{3x}} + C{e^{ - 3x}}$$
Work Step by Step
$$\eqalign{
& y' + 3y = {e^{3x}} \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + 3y = {e^{3x}} \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = 3{\text{ and }}Q\left( x \right) = {e^{3x}} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {3dx} }} = {e^{3x}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{3x}}\left( {\frac{{dy}}{{dx}} + 3y} \right) = {e^{3x}}\left( {{e^{3x}}} \right) \cr
& {e^{3x}}\frac{{dy}}{{dx}} + 3{e^{3x}}y = {e^{6x}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{e^{3x}}y} \right] = {e^{6x}} \cr
& d\left[ {{e^{3x}}y} \right] = \left( {{e^{6x}}} \right)dx \cr
& {\text{Integrate both sides}} \cr
& {e^{3x}}y = \int {{e^{6x}}} dx \cr
& {e^{3x}}y = \frac{1}{6}{e^{6x}} + C \cr
& {\text{Solve for }}y \cr
& {e^{3x}}y = \frac{1}{{6{e^{3x}}}}{e^{6x}} + \frac{C}{{{e^{3x}}}} \cr
& {e^{3x}}y = \frac{1}{6}{e^{3x}} + C{e^{ - 3x}} \cr} $$
