 Solution

The original PDF has forcing $e^{-|x|}$; the supplied TeX loses the absolute-value bars. Set $q=|\omega |$. First assume $q>0$ and $q\ne 1$. The Fourier transform of a derivative, obtained by integration by parts with the boundary terms zero, is $\widetilde{u^{\prime \prime }}=-k^2\tilde{u}$. Applying it and the preceding transform givesUsing the inverse Fourier transform pairs from part (i),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 $A{e}^{-qx}$ on the positive half-line and $B{e}^{qx}$ on the negative half-line. Continuity gives $A=B$ and derivative continuity gives $-qA=qB$, forcing both to be zero.

The PDF's condition $\omega \ne \pm 1$ also allows $\omega =0$, but no decaying solution exists when $\omega =0$. Indeed on $x>0$, $-u^{\prime \prime }=e^{-x}$ 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 $\omega \ne 0$ 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.