= Solution
Substitution gives
$$
\widetilde y(\omega)=\frac{-i}{(\omega-2i)(\omega-i-2)(\omega-i+2)}.
$$
The inverse <Fourier transform> is $(2\pi)^{-1}\int_{\mathbb R}\widetilde y(\omega)e^{i\omega t}d\omega$. For $t<0$, close clockwise in the lower half-plane, where there are no poles; hence $y(t)=0$.
For $t>0$, close counterclockwise in the upper half-plane. The three simple-pole <residues> of the full inversion integrand are
$$
\begin{array}{c|c}
\omega&\operatorname{Res}(\widetilde y e^{i\omega t})\\\hline
2i&ie^{-2t}/5\\
2+i&(1-2i)e^{(-1+2i)t}/20\\
-2+i&(-1-2i)e^{(-1-2i)t}/20
\end{array}.
$$
Multiplying their sum by $i$ and combining the conjugate exponential contributions yields
$$
\boxed{y(t)=\begin{cases}0,&t\leq0,\\[2pt]\dfrac1{10}\left[e^{-t}(2\cos2t-\sin2t)-2e^{-2t}\right],&t>0.\end{cases}}
$$
The large arcs vanish because the rational factor is $O(\omega^{-3})$ and the exponential is bounded on the chosen half-plane. At zero, the right-hand value and first derivative are both zero, matching the negative-time branch, so no unwanted delta or delta derivative appears in the second-order equation.
For an independent dynamical check, this is the <exponential forcing of an underdamped oscillator> with damping $\gamma=1$, natural frequency $\sqrt5$, damped frequency $2$, exponent $\beta=2$ and force amplitude $-1$. For positive time its substitution gives $y''+2y'+5y=-e^{-2t}$.
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