= Solution
Write $\tau=t-t'$, $y=x-x'$ and $G(y,\tau)=(4\pi D\tau)^{-1/2}e^{-y^2/(4D\tau)}$ for $\tau>0$, with $D>0$. Direct differentiation gives
$$
\partial_\tau G=\left(-\frac1{2\tau}+\frac{y^2}{4D\tau^2}\right)G
=D\partial_y^2G.
$$
Also $\int_{\mathbb R}G\,dy=1$ and, for a smooth bounded <test function> $f$,
$$
\int_{\mathbb R}G(y,\tau)f(y)\,dy
=\frac1{\sqrt\pi}\int_{\mathbb R}e^{-u^2}f(\sqrt{4D\tau}\,u)\,du
\longrightarrow f(0).
$$
Thus $G(\cdot,\tau)$ tends to a <Dirac delta distribution> as $\tau\downarrow0$. Defining the retarded kernel $K=\Theta(\tau)G$ supplies the initial impulse:
$$
\boxed{(\partial_t-D\partial_x^2)K
=\delta(t-t')\delta(x-x'),\qquad K=0\text{ for }t<t'.}
$$
The source term arises from the jump at $\tau=0$ in the distributional sense; differentiating only the smooth positive-time Gaussian would miss it.
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