= Solution
The <differential operator> on the left is $(D-3)^2$, where $D=d/dt$. Its repeated <characteristic root> suggests writing $y=e^{3t}v$. By the <product rule>,
$$
y'=e^{3t}(v'+3v),\qquad y''=e^{3t}(v''+6v'+9v),
$$
so the <ordinary differential equation> reduces to $v''=3(t+1)$. Two <antiderivatives> give
$$
v'=\frac32t^2+3t+A,\qquad
v=\frac12t^3+\frac32t^2+At+B.
$$
The <boundary conditions> imply $v(0)=1$ and $v(1)=0$, hence $B=1$ and $A=-3$. Therefore
$$
\boxed{y(t)=e^{3t}\left(\frac12t^3+\frac32t^2-3t+1\right)}.
$$
The calculation of $v''$ verifies the <ordinary differential equation>, and the bracket takes the values one and zero at the two endpoints. These <boundary conditions> determine both constants uniquely.
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