= Solution
Factor the cubic denominator as $(\omega-i)(\omega^2+4)$. Substitution gives
$$
\widetilde y(\omega)=\frac{i\omega}{(\omega-i)^2(\omega^2+4)}.
$$
There is a double <pole> at $i$ and simple poles at $2i,-2i$. For $t>0$, close the inverse-transform contour counterclockwise in the upper half-plane. The <residue> at the double pole is
$$
\begin{aligned}
\operatorname{Res}_{\omega=i}\big(\widetilde y e^{i\omega t}\big)
&=\left.\frac d{d\omega}\left(\frac{i\omega e^{i\omega t}}{\omega^2+4}\right)\right|_{\omega=i}\\
&=i\left(\frac59-\frac t3\right)e^{-t},
\end{aligned}
$$
and the residue at $2i$ is $-ie^{-2t}/2$. Multiplication by $i$ from the <residue theorem> therefore gives $y(t)=(t/3-5/9)e^{-t}+e^{-2t}/2$.
For $t<0$, the clockwise lower closure encloses only $-2i$, with residue $-ie^{2t}/18$. The orientation supplies $-i$, giving $y(t)=-e^{2t}/18$. Hence the continuous solution is
$$
\boxed{y(t)=\begin{cases}-\dfrac1{18}e^{2t},&t<0,\\[2pt]-\dfrac1{18},&t=0,\\[2pt]\left(\dfrac t3-\dfrac59\right)e^{-t}+\dfrac12e^{-2t},&t>0.\end{cases}}
$$
The large inverse-transform arcs vanish: the rational factor is $O(\omega^{-3})$, while the exponential is bounded on the appropriate closure.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2-2016-ib-fourier-solution.png]
{title=The two-sided decaying Fourier solution; it and its first derivative are continuous at zero}
The graph tends to zero from below as $t\to-\infty$, passes through $y(0)=-1/18$ with slope $-1/9$, reaches a small negative minimum after zero, then crosses the axis, reaches a positive maximum and tends to zero from above. The positive-time stationary points solve $8/9-t/3=e^{-t}$, and the crossing solves $t/3-5/9+e^{-t}/2=0$; these equations determine the sketch without guessing signs.
For a distributional check, $y$ and $y'$ are continuous, while $y''(0^+)-y''(0^-)=1$. The <jump conditions for a third-order Green function> therefore produce the delta in $f'=\delta-e^{-t}H(t)$, and each smooth branch satisfies the remaining equation. The nonzero solution for $t<0$ is consistent with the stipulated decay at both temporal infinities. Those are two-sided conditions, not a causality assumption: this is the setting of a <decaying two-sided Green function of a constant-coefficient ordinary differential equation> and <causality versus stability of a Green function>.
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