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