I am working through some exercises in the text Integral Transforms and Their Applications by Debnath and Bhatta. This exercise comes in Chapter 4 - Applications of Laplace Transforms.
Solve the telegraph equation $$ u_{tt} - c^2u_{xx} + 2au_x = 0 $$ subject to the initial conditions $$u(x, 0) = 0, u_t(x, 0) = g(x)$$ where $-\infty < x < \infty$ and $t > 0.$
Let $\mathcal{L}u(x, t) = U(x, s)$ (that is, take the Laplace transform in the time variable). Then after taking transforms the equation becomes $$s^2U - g(x) - c^2U_{xx} + 2asU = 0$$
which can be rearranged to
$$ ((s+a)^2 - a^2)U - c^2U_{xx} = g(x) $$
after some collecting of terms and completing the square. Solving the corresponding homogeneous problem gives the solution
$$U_c = c_1(s)e^{\lambda x} + c_2(s)e^{-\lambda x}$$
where $\lambda = \frac{1}{c} \sqrt{(s+a)^2 - a^2}.$ To obtain the particular solution, we use variation of parameters to arrive at
$$U_p = \frac{1}{2\lambda} \int_0 ^x e^{\lambda(x-t)} g(t) \ dt - \frac{1}{2\lambda} \int_0 ^x e^{-\lambda(x-t)} g(t) \ dt = \int_0 ^x g(t) \sinh(\lambda(x-t)) \ dt. $$
My issue now is that this isn't phrased as a typical boundary value problem, but I expect that the solution $u(x, t)$ should decay to zero as $x \to \pm \infty.$ I am also unsure on how to handle the unknown functions $c_1 (s)$ and $c_2 (s)$. If I assumed this decay, then both $c_1$ and $c_2$ would be the zero function, which I don't believe is correct. Is there a way to proceed with the solution as I've written it?
Following Gonçalo's suggestion, I've taken the Fourier transform of the ODE for $U(x, s)$. With $\mathcal{F}(U(x, s)) = \overline{U}(\omega, s),$ we have
$$ -\omega^2 \overline{U}(\omega, s) - \lambda^2 \overline{U}(\omega, s) = \frac{\overline{g}(\omega)}{c^2} $$
which gives
$$ \overline{U}(\omega, s) = -\frac{\overline{g}(\omega)}{c^2(\omega^2 + \lambda^2)}.$$
Using the book's convention of defining the Fourier transform with
$$ \overline{F}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty} ^\infty e^{-i\omega x} F(x) \ dx $$
we arrive at
$$ U(x, s) = -\frac{1}{\lambda c^2} \sqrt{\frac{\pi}{2}} \int_{-\infty} ^\infty e^{\lambda |y|} g(x-y) \ dy $$
which is a convolution, but now I'm still at a roadblock.