= Solution
The <laplace transform> is $\widehat f(p)=\int_0^\infty e^{-pt}f(t),dt$, and inversion is
$$
f(t)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{pt}\widehat f(p),dp,
$$
where the vertical <Bromwich contour> lies to the right of all singularities. For $t>0$ it is closed leftwards, and for $t<0$ rightwards, where the exponential decays.
Transforming the PDE and imposing boundedness as $x\to\infty$ gives
$$
U(x,p)=\frac1{p(p+1)}+\frac{p-1}{p(p+1)}e^{-x\sqrt{p/D}},
$$
with the square root cut conventionally along $(-\infty,0]$. Therefore
$$
u(x,t)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{pt}U(x,p),dp.
$$
There are poles at $p=0,-1$ (with cancellation assessed in the complete expression) and a <branch point> at $0$; close left for $t>0$ and wrap the cut while including the relevant residues.
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