Answer
$y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+\frac{1}{5}+\frac{1}{8}e^{x}$
Work Step by Step
$y''+2y'+5y=1+e^x$
$r^2+2r+5=0$
$r=-1+2i$
$y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))$
$y=A+Be^x$
$y'=Be^x$
$y''=Be^x$
$Be^x+2Be^x+5A+5Be^x=1+e^x$
$A=\frac{1}{5}$
$B=\frac{1}{8}$
$y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+\frac{1}{5}+\frac{1}{8}e^{x}$