= Solution
Insert
$$
w(z)=\int_\gamma e^{zt}f(t)\,dt
$$
into the equation. Since $w'=\int_\gamma te^{zt}f\,dt$ and $w''=\int_\gamma t^2e^{zt}f\,dt$, use $ze^{zt}=\partial_te^{zt}$ and <integration by parts>:
$$
zw''+2aw'+zw
=\left[e^{zt}(1+t^2)f(t)\right]_{\partial\gamma}
+\int_\gamma e^{zt}
\left[
2atf-\frac d{dt}\bigl((1+t^2)f\bigr)
\right]dt.
$$
The integral vanishes when
$$
(1+t^2)f'=2(a-1)tf,
$$
so, up to a constant and on a consistently chosen branch,
$$
\boxed{\ f(t)=(1+t^2)^{a-1}\ }.
$$
The remaining contour condition is
$$
\boxed{\
\left[e^{zt}(1+t^2)^a\right]_{\partial\gamma}=0\
}.
$$
This is the <Laplace contour-integral method for a linear differential equation>.
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