= Solution
The <characteristic polynomial> factors as $(r+1)(r+2)$, so the <homogeneous solution> is $Ae^{-x}+Be^{-2x}$. The forcing $e^{-x}$ already occurs in that solution, so a constant multiple of $e^{-x}$ cannot be a <particular solution>. Use $y_p=Cxe^{-x}$ instead. Its <derivatives> are $C(1-x)e^{-x}$ and $C(x-2)e^{-x}$, so substitution gives $Ce^{-x}$ and fixes $C=1$. Thus
$$
\boxed{y=Ae^{-x}+Be^{-2x}+xe^{-x}.}
$$
The extra factor of $x$ accounts for the repeated forcing exponent in the <method of undetermined coefficients>.
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