= Solution
Substituting
$$
y(z)=\int_\gamma e^{zt}f(t)\,dt,
\qquad y^{(3)}(z)=\int_\gamma e^{zt}t^3f(t)\,dt,
$$
and using $ze^{zt}=\partial_te^{zt}$ gives
$$
zy^{(3)}+2y
=[e^{zt}t^3f(t)]_{\partial\gamma}
+\int_\gamma e^{zt}\{2f(t)-(t^3f(t))'\}\,dt.
$$
It is therefore enough to require
$$
(t^3f)'=2f
$$
and choose $\gamma$ so that the <integral> converges and $e^{zt}t^3f(t)$ vanishes at its endpoints. Solving the amplitude equation,
$$
\boxed{\frac{f'}f=\frac2{t^3}-\frac3t,
\qquad
f(t)=Ct^{-3}e^{-1/t^2}.}
$$
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