Answer
$y=c_{1}e^{-x}cos(2x)+c_{2}e^{-x}sin(2x)+\frac{1}{5}+\frac{1}{8}e^{x}$
Work Step by Step
$y''+2y'+5y=0$
Use auxiliary equation
$r^{2}+2r+5=0$
$r=\frac{-2±\sqrt (4-20)}{2(1)}$
$r=\frac{-2±4i}{2}$
$r=-1±2i$
$α=-1$
$β=2$
$y_{c}=e^{αx}(c_{1}cosβx+c_{2}sinβx)$
$y_{c}=e^{-x}(c_{1}cos2x+c_{2}sin2x)$
Given:
Since $G(x)=1+e^{x}$ we seek a particular solution of form $y_{p}=A+Be^{x}$
$y_{p}'=Be^{x}$
$y_{p}''=Be^{x}$
Plug into original equation
$Be^{x}+2Be^{x}+5A+5Be^{x}=1+e^{x}$
$Be^{x}(1+2+5)+5A=1+e^{x}$
$8Be^{x}+5A=1+e^{x}$
$B=\frac{1}{8}$
$A=\frac{1}{5}$
$y_{p}=\frac{1}{5}+\frac{1}{8}e^{x}$
$y=y_{p}+y_{c}$
$y=c_{1}e^{-x}cos(2x)+c_{2}e^{-x}sin(2x)+\frac{1}{5}+\frac{1}{8}e^{x}$
