Answer
$$y = {e^{5x}} + \frac{{24}}{{{e^{4x}}}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} + 4y = 9{e^{5x}},\,\,\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 25 \cr
& {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{comparing }}\frac{{dy}}{{dx}} + 4y = 9{e^{5x}}{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is 4 }} \cr
& {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int {4dx} }} = {e^{4x}} \cr
& {\text{multiplying both sides of the differential equation by }}{e^x} \cr
& {e^{4x}}\frac{{dy}}{{dx}} + 4{e^{4x}}y = 9{e^{9x}} \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {{e^{4x}}y} \right] = 9{e^{9x}} \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& {e^{4x}}y = \int {9{e^{9x}}} dx \cr
& {e^{4x}}y = {e^{9x}} + C \cr
& y = {e^{5x}} + \frac{C}{{{e^{4x}}}}\,\,\,\,\left( {\bf{1}} \right) \cr
& {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 0 \right) = 50 \cr
& {\text{substituting these values into }}\left( {\bf{1}} \right) \cr
& 25 = {e^{5\left( 0 \right)}} + \frac{C}{{{e^{0\left( 0 \right)}}}} \cr
& C = 24 \cr
& \cr
& {\text{substitute }}C = 24{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& y = {e^{5x}} + \frac{{24}}{{{e^{4x}}}} \cr} $$
