Answer
$$y = 2{e^x} + 48{e^{ - x}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} + y = 4{e^x},\,\,\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 50 \cr
& {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{comparing }}\frac{{dy}}{{dx}} + y = 4{e^x}{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is 1 }} \cr
& {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int {dx} }} = {e^x} \cr
& {\text{multiplying both sides of the differential equation by }}{e^x} \cr
& {e^x}\frac{{dy}}{{dx}} + {e^x}y = 4{e^{2x}} \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {{e^x}y} \right] = 4{e^{2x}} \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& {e^x}y = \int {4{e^{2x}}} dx \cr
& {e^x}y = 4\left( {\frac{{{e^{2x}}}}{2}} \right) + C \cr
& {e^x}y = 2{e^{2x}} + C \cr
& y = 2{e^x} + \frac{C}{{{e^x}}}\,\,\,\,\left( {\bf{1}} \right) \cr
& {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 0 \right) = 50 \cr
& {\text{substituting these values into }}\left( {\bf{1}} \right) \cr
& 50 = 2{e^0} + \frac{C}{{{e^0}}} \cr
& C = 48 \cr
& \cr
& {\text{substitute }}C = 48{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& y = 2{e^x} + \frac{{48}}{{{e^x}}} \cr
& or \cr
& y = 2{e^x} + 48{e^{ - x}} \cr} $$