= Solution
Substitute the <Laplace contour-integral method for a linear differential equation> into the <differential equation> and use $z e^{zt}=\partial_t e^{zt}$. <Integration by parts> around a closed contour avoiding singularities gives
$$
0=\int_C e^{zt}\{-[(1+t^2)f(t)]'-2tf(t)\}\,dt.
$$
It suffices to solve $(1+t^2)f'+4tf=0$, giving
$$
\boxed{f(t)=\frac{C_0}{(1+t^2)^2}.}
$$
Nonzero contour integrals arise by enclosing one or both double poles $t=\pm i$. At $t=i$, the <residue> is
$$
\operatorname{Res}_{t=i}\frac{e^{zt}}{(1+t^2)^2}=-\frac14(z+i)e^{iz}.
$$
The conjugate pole gives the conjugate expression for real $z$. Taking real and imaginary linear combinations yields the real solutions
$$
\boxed{y_1(z)=\sin z-z\cos z,\qquad y_2(z)=\cos z+z\sin z.}
$$
Their <derivatives> are $y_1'=z\sin z$, $y_2'=z\cos z$, and their <Wronskian> is $-z^2$, so they are <linearly independent> on every interval avoiding zero. Both extend analytically across zero and remain independent as functions; the vanishing <Wronskian> at zero reflects the singular leading coefficient of the printed <differential equation> there.
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