Answer
$$y = \frac{1}{{{e^{2x}}}} + \frac{C}{{{e^{3x}}}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} + 3y = {e^{ - 2x}} \cr
& {\text{The equation is in the form }}\frac{{dy}}{{dp}} + p\left( x \right)y = q\left( x \right) \cr
& {\text{with }}p\left( x \right) = 3{\text{ and }}q\left( x \right) = {e^{ - 2x}} \cr
& \cr
& {\text{Use the method of the integrating factors}}{\text{, calculate }}\mu = {e^{\int {p\left( x \right)dx} }} \cr
& \mu = {e^{\int {p\left( x \right)dx} }} = {e^{\int 3 dx}} \cr
& \mu = {e^{3x}} \cr
& \cr
& {\text{Multiply both sides of the equation by }}\mu \cr
& {e^{3x}}\frac{{dy}}{{dx}} + 3{e^{3x}}y = {e^{ - 2x}}{e^{3x}} \cr
& {e^{3x}}\frac{{dy}}{{dx}} + 3{e^{3x}}y = {e^x} \cr
& {\text{Express the result as }}\frac{d}{{dx}}\left[ {\mu y} \right] = {e^x}{\text{. Then}}{\text{,}} \cr
& \frac{d}{{dx}}\left[ {{e^{3x}}y} \right] = {e^x} \cr
& {\text{Integrate both sides}} \cr
& {e^{3x}}y = \int {{e^x}} dx \cr
& {e^{3x}}y = {e^x} + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{{e^x}}}{{{e^{3x}}}} + \frac{C}{{{e^{3x}}}} \cr
& y = \frac{1}{{{e^{2x}}}} + \frac{C}{{{e^{3x}}}} \cr} $$
