= Solution
The <differential operator> is $(D-1)^2$, with $D=d/dt$. Because $e^t$ already occurs in the <complementary function> with a repeated <root of a polynomial>, an unmodified trial $Ce^t$ gives zero. Put $f_p=e^t h(t)$. The <product rule> yields
$$
(D-1)(e^th)=e^th',\qquad (D-1)^2(e^th)=e^th''.
$$
Thus $h''=2$, and we may choose $h=t^2$, giving
$$
\boxed{f_p(t)=t^2e^t.}
$$
The other terms arising on twice integrating $h''=2$ give $(C_1+C_2t)e^t$, exactly the <complementary function>, so they need not be included in the <particular integral>.
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