= Solution
The initial profile is $f(x)=e^{-|x|}$, with the absolute value confirmed in the original PDF. Its <Fourier transform> is
$$
\widetilde f(k)=\int_0^\infty e^{-x}(e^{-ikx}+e^{ikx})dx=\frac{2}{1+k^2}.
$$
Using $\widetilde u_t(k,0)=0$ in the regular cosine-sine form gives
$$
\widetilde u(k,t)=\frac{2e^{-\gamma t}}{1+k^2}\left[\cos(ckt)+\gamma\frac{\sin(ckt)}{ck}\right],
$$
with the quotient interpreted as $t$ when $ck=0$. Put $v_0=|c|>0$. Inverting, or applying the <D'Alembert formula> to $e^{\gamma t}u$, whose initial velocity is $\gamma f$, gives
$$
\boxed{u(x,t)=\frac{e^{-\gamma t}}2\left[e^{-|x-v_0t|}+e^{-|x+v_0t|}+\frac{\gamma}{v_0}\int_{x-v_0t}^{x+v_0t}e^{-|s|}ds\right].}
$$
For an explicit expression at $t\geq0$, let $r=|x|$ and $s=v_0t$. The integral is $2e^{-r}\sinh s$ for $r\geq s$ and $2(1-e^{-s}\cosh r)$ for $r\leq s$. Therefore
$$
\boxed{u(x,t)=e^{-\gamma t}\begin{cases}
e^{-r}\left[\cosh(v_0t)+\dfrac{\gamma}{v_0}\sinh(v_0t)\right],&r\geq v_0t,\\[3pt]
e^{-v_0t}\cosh r+\dfrac{\gamma}{v_0}\left[1-e^{-v_0t}\cosh r\right],&r\leq v_0t.
\end{cases}}
$$
The branches agree at $r=v_0t$. At zero time, the integral vanishes and the translated profiles coincide, recovering $f$. Their time derivatives cancel, while the integral contributes $\gamma f$, cancelling the derivative of the prefactor, so $u_t(x,0)=0$.
For $c=0$ the limiting answer is $u=(1+\gamma t)e^{-\gamma t}e^{-|x|}$. The cusp in $f$ means the solution is understood as a <weak solution>, classically smooth away from the characteristic fronts $x=\pm v_0t$; it is not globally twice differentiable. No assumption that $\gamma$ is positive is needed for this initial-value solution.
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