= Solution
The original PDF has forcing $e^{-|x|}$; the supplied TeX loses the absolute-value bars. Set $q=|\omega|$. First assume $q>0$ and $q\ne1$. The <Fourier transform of a derivative>, obtained by <integration by parts> with the boundary terms zero, is $\widetilde{u''}=-k^2\widetilde u$. Applying it and the preceding transform gives
$$
\boxed{\widetilde u(k)=\frac2{(k^2+1)(k^2+q^2)}}=\frac2{q^2-1}\left(\frac1{k^2+1}-\frac1{k^2+q^2}\right).
$$
Using the inverse <Fourier transform> pairs from part (i),
$$
\boxed{u(x)=\frac{e^{-|x|}-q^{-1}e^{-q|x|}}{q^2-1},\qquad q=|\omega|>0,\ q\ne1}.
$$
Both decay conditions hold. The first derivatives of the two exponential terms cancel at zero, so $u$ is continuously differentiable there; the second derivatives have matching limits and the <differential equation> holds at zero as well as on both half-lines. For uniqueness, a difference solves the homogeneous equation. Decay gives $Ae^{-qx}$ on the positive half-line and $Be^{qx}$ on the negative half-line. Continuity gives $A=B$ and derivative continuity gives $-qA=qB$, forcing both to be zero.
The PDF's condition $\omega\ne\pm1$ also allows $\omega=0$, but \b[no decaying solution exists when $\omega=0$]. Indeed on $x>0$, $-u''=e^{-x}$ and decay forces $u=-e^{-x}$; on $x<0$, decay forces $u=-e^x$. Their one-sided derivatives at zero are $1$ and $-1$, which cannot match. The missing condition $\omega\ne0$ is therefore a genuine source omission, not a transcription repair. The transform expression at $q=0$ has a nonintegrable $k^{-2}$ singularity, consistently with this obstruction.
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