= Solution
Substitution of
$$
w(z)=\int_\gamma e^{zt}f(t)\,dt
$$
and integration by parts give
$$
w'''-zw
=\int_\gamma e^{zt}(t^3f+f')\,dt
-\left[e^{zt}f(t)\right]_{\partial\gamma}.
$$
Thus take
$$
\boxed{f(t)=e^{-t^4/4}},
$$
and require the endpoint term to vanish. Infinite contour ends must lie in sectors where $\operatorname{Re}(t^4)>0$, namely sectors centred on the positive and negative real and imaginary axes, each of angular width $\pi/4$.
Let the common final segment be the positive real ray from $0$ to $+\infty$, and choose the initial rays from $-\infty$, $+i\infty$, and $-i\infty$ to $0$. Call the resulting contours $\gamma_1,\gamma_2,\gamma_3$. Put
$$
A_r=\int_0^\infty t^re^{-t^4/4}\,dt
=4^{(r-3)/4}\Gamma\!\left(\frac{r+1}{4}\right).
$$
Since $w_i^{(r)}(0)=\int_{\gamma_i}t^re^{-t^4/4}\,dt$, for $r=0,1,2$ the three rows are
$$
\begin{pmatrix}
2A_0&0&2A_2\\
(1-i)A_0&2A_1&(1+i)A_2\\
(1+i)A_0&2A_1&(1-i)A_2
\end{pmatrix}.
$$
Consequently
$$
\det\begin{pmatrix}
w_1(0)&w_1'(0)&w_1''(0)\\
w_2(0)&w_2'(0)&w_2''(0)\\
w_3(0)&w_3'(0)&w_3''(0)
\end{pmatrix}
=-16iA_0A_1A_2.
$$
Using the <Gamma function> reflection identity,
$$
\boxed{\det=-2\sqrt2\,i\,\pi^{3/2}\ne0}.
$$
The initial-data vectors are therefore <linear independence>[linearly independent], so the three contour integrals are linearly independent solutions.
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