Answer
The general solution is of the form: $y= c_1e^{3x}+c_2e^{-2x}-\frac{1}{5}xe^{-2x}-\frac{1}{6}$
Work Step by Step
The characteristic equation for $y''-y'-6y=1+e^{-2x}$ is $t^2-t-6=0$ with solutions $t=-2$ and $t=3$
For particular solution, we have $Ae^{-2x}+B$ gives $A=-1/5$ and $B=-1/6$
The general solution is of the form: $y= c_1e^{3x}+c_2e^{-2x}-\frac{1}{5}xe^{-2x}-\frac{1}{6}$
