= Solution
The <Laplace transform> is $\overline f(p)=\int_0^\infty e^{-pt}f(t)dt$ wherever it converges. For sufficiently large $\operatorname{Re}p$, suitable exponential bounds justify <integration by parts> and <differentiation under the integral sign>, giving
$$
\boxed{\mathcal L[f'](p)=p\overline f(p)-f(0^+),\qquad\mathcal L[tf](p)=-\overline f'(p).}
$$
For a nonnegative <integer> $n$ and real $p>0$, substitute $u=pt$ in the <gamma integral>. The result extends holomorphically to $\operatorname{Re}p>0$, giving
$$
\boxed{\mathcal L[t^n](p)=\frac{n!}{p^{n+1}},\qquad\operatorname{Re}p>0.}
$$
This also follows recursively from the second identity, starting with $\mathcal L[1]=1/p$.
Put $G=\overline g$, $g_0=g(0^+)$ and $g_1=g'(0^+)$. For a solution with finite initial values and the required transform bounds, the transformed equation is
$$
-\frac d{dp}(p^2G-pg_0-g_1)+(pG-g_0)+\frac d{dp}(pG-g_0)+2G=0.
$$
The initial-value terms cancel, leaving $(p-p^2)G'+(3-p)G=0$. On the real interval $p>1$, a nonzero solution satisfies
$$
\frac{G'}G=\frac{3-p}{p^2-p}=-\frac3p+\frac2{p-1}.
$$
Integration gives $G=C(p-1)^2/p^3=C(1/p-2/p^2+1/p^3)$. Inverting term by term yields
$$
\boxed{g(t)=C\left(1-2t+\frac{t^2}2\right).}
$$
Taking $C=1$ gives a nonzero solution, verified by direct substitution. It is $L_2(t)$, and the rational transform agrees with the general <Laplace transform of a Laguerre polynomial>. No claim about solutions with singular initial behavior is needed to produce this regular solution.
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