Answer
$y=c_{1}cos(\frac{1}{3}x)+c_{2}sin(\frac{1}{3}x)+\frac{1}{37}e^{2x}$
Work Step by Step
$9y''+y=e^{2x}$
$9r^{2}+1=0$
$r=i\sqrt{\frac{1}{9}}=\frac{1}{3}i$
$y=c_{1}cos(\frac{1}{3}x)+c_{2}sin(\frac{1}{3}x)$
$y_{p}=Ae^{2x}$
$y_{p}'=2Ae^{2x}$
$y_{p}''=4Ae^{2x}$
$9(4Ae^{2x})+Ae^{2x}=36Ae^{2x}+Ae^{2x}=37Ae^{2x}=e^{2x}$
$A=\frac{1}{37}$
$y=c_{1}cos(\frac{1}{3}x)+c_{2}sin(\frac{1}{3}x)+\frac{1}{37}e^{2x}$
