= Solution
Use the <method of undetermined coefficients> with the <polynomial> $f_p=at^3+bt^2+ct+d$. Substitution into $f_p''+3f_p'+2f_p$ gives
$$
2at^3+(9a+2b)t^2+(6a+6b+2c)t+(2b+3c+2d).
$$
Matching successive coefficients with the forcing gives $a=1$, $b=-3$, $c=4$ and $d=1$. Hence a <particular integral> is
$$
\boxed{f_p(t)=t^3-3t^2+4t+1.}
$$
Indeed $f_p'=3t^2-6t+4$ and $f_p''=6t-6$ give $f_p''+3f_p'+2f_p=2t^3+3t^2-4t+8$. The <complementary function>, if wanted, is $C_1e^{-t}+C_2e^{-2t}$.
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