= Solution
Apply the time <Laplace transform> to the <Airy equation>. Extend $g$ past $T$ in any suitable way and write
$$
U(x,p)=\int_0^\infty e^{-pt}u(x,t)dt,\qquad G(p)=\int_0^\infty e^{-pt}g(t)dt.
$$
For example, setting $g=0$ after $T$ makes $G(p)=\int_0^T e^{-pt}g(t)dt$; the final solution for $t<T$ is independent of this extension. The <Laplace transform of a derivative> gives
$$
U_{xxx}+pU=u_0(x),\qquad U_x(0,p)=G(p).
$$
Take the <principal cube root> $\lambda=p^{1/3}$ for $\operatorname{Re}p>0$. The three <characteristic roots> of the spatial <ordinary differential equation> are
$$
-\lambda,\qquad r_1=\lambda e^{i\pi/3},\qquad r_2=\lambda e^{-i\pi/3}.
$$
Since $|\arg\lambda|<\pi/6$, only $-\lambda$ has negative <real part>. Thus spatial decay leaves just one homogeneous exponential, which the single prescribed <Neumann boundary condition> determines.
The <Airy resolvent kernel> on the whole real line is
$$
R_p(s)=\begin{cases}
\displaystyle\frac{e^{-\lambda s}}{3\lambda^2},&s\geq0,\\
\displaystyle-\frac{e^{r_1s}}{3r_1^2}-\frac{e^{r_2s}}{3r_2^2},&s<0.
\end{cases}
$$
It decays at both ends, is continuous together with its first <derivative>, and satisfies $R_p''(0+)-R_p''(0-)=1$. Consequently $(\partial_s^3+p)R_p=\delta_0$ as a <distribution>. A particular solution is the <Green-function representation>
$$
V(x,p)=\int_0^\infty R_p(x-y)u_0(y)\,dy.
$$
Adding the decaying homogeneous mode to impose the boundary derivative gives
$$
U(x,p)=V(x,p)+\frac{e^{-\lambda x}}\lambda\left[V_x(0,p)-G(p)\right],
\qquad
V_x(0,p)=\int_0^\infty R_p'(-y)u_0(y)\,dy.
$$
Every term is known. If $\widetilde u_0(r)=\int_0^\infty e^{-ry}u_0(y)dy$ denotes the spatial <Laplace transform>, then
$$
V_x(0,p)=-\frac13\left[\frac{\widetilde u_0(r_1)}{r_1}+\frac{\widetilde u_0(r_2)}{r_2}\right].
$$
The <Bromwich inversion formula> now gives the required <integral representation>:
$$
\boxed{u(x,t)=\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{pt}\left\{\int_0^\infty R_p(x-y)u_0(y)\,dy+\frac{e^{-p^{1/3}x}}{p^{1/3}}\left[\int_0^\infty R_p'(-y)u_0(y)\,dy-G(p)\right]\right\}dp.}
$$
Here $\gamma>0$ is to the right of any singularities required by the growth of the data. The usual decay or growth hypotheses are understood for this <Laplace transform> construction; when absolute inversion is unavailable, the vertical integral is interpreted as the limit of truncated <Bromwich contours>. The transformed <ordinary differential equation> verifies the <partial differential equation> and <initial condition>, while differentiating at $x=0$ gives exactly $G(p)$ and hence $u_x(0,t)=g(t)$. The derivative compatibility $u_0'(0)=g(0)$ makes the two data agree at the corner.
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