Answer
$$y = \frac{1}{{3{e^x}}} + \frac{C}{{{e^{4x}}}}$$
Work Step by Step
$$\eqalign{
& {e^x}y' + 4{e^x}y = 1 \cr
& {\text{Divide the equation by }}{e^x} \cr
& \frac{{{e^x}y'}}{{{e^x}}} + \frac{{4{e^x}y}}{{{e^x}}} = \frac{1}{{{e^x}}} \cr
& y' + 4y = {e^{ - x}} \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + 4y = {e^{ - x}} \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = 4{\text{, }}Q\left( x \right) = {e^{ - x}} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {4dx} }} = {e^{4x}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{4x}}\left( {\frac{{dy}}{{dx}} + 4y} \right) = {e^{ - x}}{e^{4x}} \cr
& {e^{4x}}\frac{{dy}}{{dx}} + 4{e^{4x}}y = {e^{3x}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{e^{4x}}y} \right] = {e^{3x}} \cr
& d\left[ {{e^{4x}}y} \right] = {e^{3x}}dx \cr
& {\text{Integrate both sides}} \cr
& {e^{4x}}y = \int {{e^{3x}}} dx \cr
& {e^{4x}}y = \frac{1}{3}{e^{3x}} + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{{e^{3x}}}}{{3{e^{4x}}}} + \frac{C}{{{e^{4x}}}} \cr
& y = \frac{1}{{3{e^x}}} + \frac{C}{{{e^{4x}}}} \cr} $$
