Answer
$y=c_{1}cos(3x)+c_{2}sin(3x)+\frac{1}{13}e^{-2x}$
Work Step by Step
$y''+9y=e^{-2x}$
Use auxiliary equation
$r^{2}+9=0$
$r^{2}=-9$
$r=±3i$
$α=0$
$β=3$
$y_{c}=e^{αx}(c_{1}cos(βx)+c_{2}sin(βx))$
$y_{c}=e^{0x}(c_{1}cos(3x)+c_{2}sin(3x))$
$y_{c}=c_{1}cos(3x)+c_{2}sin(3x)$
$y_{p}=Ae^{-2x}$
$y_{p}'=-2Ae^{-2x}$
$y_{p}''=4Ae^{-2x}$
Plug into original equation
$4Ae^{-2x}+9Ae^{-2x}=e^{-2x}$
$4A+9A=1$
$A=\frac{1}{13}$
$y_{p}=\frac{1}{13}e^{-2x}$
$y=y_{c}+y_{p}$
$y=c_{1}cos(3x)+c_{2}sin(3x)+\frac{1}{13}e^{-2x}$
