Answer
$c_1e^{3x}+c_2e^{-2x}-\frac{1}{5}xe^{-2x}-\frac{1}{6}$
Work Step by Step
$y''-y'-6y=1+e^{-2x}$ which has the characteristic equation $t^2-t-6=0$ with solutions $t=-2$ and $t=3$
Particular solution: $Ae^{-2x}+B$ giving $A=-1/5$ and $B=-1/6$
General solution: $c_1e^{3x}+c_2e^{-2x}-\frac{1}{5}xe^{-2x}-\frac{1}{6}$
