= Solution
Take the <Fourier transform> convention $\widetilde y(\omega)=\int_{\mathbb R}y(t)e^{-i\omega t}\,dt$, with inverse factor $e^{i\omega t}/(2\pi)$. Then $\widetilde{y^{(n)}}=(i\omega)^n\widetilde y$, classically when the boundary terms vanish or in the sense of <tempered distributions>.
Multiply the given frequency relation by its denominator and then by $-i$:
$$
(-i\omega^3-\omega^2-4i\omega-4)\widetilde y=i\omega\widetilde f.
$$
Recognizing the four derivative multipliers gives
$$
\boxed{y'''+y''-4y'-4y=f',\qquad(D+1)(D-2)(D+2)y=f',\quad D=\frac d{dt}.}
$$
The convention is stated explicitly because reversing the Fourier exponential changes intermediate frequency signs. The distributional form will matter at the jump of $f$ in part (b).
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