Answer
$$y = \frac{1}{{10}}{e^{5x}} + \frac{{29}}{{10}}{e^{ - 5x}}$$
Work Step by Step
$$\eqalign{
& y' + 5y = {e^{5x}},{\text{ }}y\left( 0 \right) = 3 \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + 5y = {e^{5x}} \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = 5{\text{, }}Q\left( x \right) = {e^{5x}} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {5dx} }} = {e^{5x}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{5x}}\frac{{dy}}{{dx}} + 5{e^{5x}}y = {e^{5x}}{e^{5x}} \cr
& {e^{5x}}\frac{{dy}}{{dx}} + 5{e^{5x}}y = {e^{10x}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{e^{5x}}y} \right] = {e^{10x}} \cr
& d\left[ {{e^{5x}}y} \right] = {e^{10x}}dx \cr
& {\text{Integrate both sides}} \cr
& {e^{5x}}y = \int {{e^{10x}}} dx \cr
& {e^{5x}}y = \frac{1}{{10}}{e^{10x}} + C \cr
& {\text{Solve for }}y \cr
& y = \frac{1}{{10}}\frac{{{e^{10x}}}}{{{e^{5x}}}} + \frac{C}{{{e^{5x}}}} \cr
& y = \frac{1}{{10}}{e^{5x}} + \frac{C}{{{e^{5x}}}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 0 \right) = 3 \cr
& 3 = \frac{1}{{10}}{e^0} + \frac{C}{{{e^0}}} \cr
& C = \frac{{29}}{{10}} \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{1}{{10}}{e^{5x}} + \frac{{29}}{{10}}{e^{ - 5x}} \cr} $$
