 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 $C{e}^t$ gives zero. Put $f_p=e^{t}h(t)$. The product rule yieldsThus $h^{\prime \prime }=2$, and we may choose $h=t^2$, givingThe other terms arising on twice integrating $h^{\prime \prime }=2$ give $(C_1+C_{2}t)e^t$, exactly the complementary function, so they need not be included in the particular integral.