= Solution
The <linear ordinary differential equation> has differential operator $(D-3)^2$. Put $y=e^{3x}v$ and use the <product rule> twice:
$$
y'=e^{3x}(3v+v'),\qquad
y''=e^{3x}(9v+6v'+v'').
$$
Substitution cancels the $v$ and $v'$ terms, leaving $v''=4$. Two <integrations> give $v=2x^2+C_1x+C_2$. Hence the <general solution> is
$$
\boxed{y(x)=e^{3x}(2x^2+C_1x+C_2),\qquad C_1,C_2\in\mathbb R.}
$$
The two arbitrary constants account for the order of the <differential equation>; the $x^2$ factor in the particular term reflects the <repeated root> $3$ of the <characteristic polynomial>.
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