PDE : Solving the damped (nonhomogeneous) heat conduction equation $\frac{\partial v(x,t)}{\partial t}- C\frac{\partial^2 v(x,t)}{\partial x^2}=b{v(x,t)}$
$IC: v(x,0) = f(x)$
$ -\infty \le x  \le \infty ;$ b and C are both constants; $C > 0$
A particular solution would be 
$v_p(x,t) = c_p e^{ikx-\alpha t}$ 
which as I understand it would solve the homogenous version of the initial PDE:
$\frac{\partial v(x,t)}{\partial t}- C\frac{\partial^2 v(x,t)}{\partial x^2}=0$
Now how would one procede to solve this taking into account for the nonhomogenous version combining with the particular solution and superposition? Help is appreciated.
 A: The idea is to diagonalize the operator $-\frac{d^{2}}{dx^{2}}$ and then to expand $v(x,t)$ in the eigenfunctions of this operator to build the final solution:
$$
           v(x,t)=\int_{-\infty}^{\infty}C(s,t)e^{isx}ds \\
           \frac{\partial v}{\partial t}-C\frac{\partial^{2}v}{\partial x^{2}} = bv(x,t) \\
         \int_{-\infty}^{\infty}\left(\frac{\partial C}{\partial t}+s^{2}C(s,t)-bC\right)e^{-isx}ds=0
$$
So the coefficient function $C$ must satisfy
$$
                  \frac{\partial C}{\partial t}=-s^{2}C+bC\\
                   C(s,t) = K(s)e^{-s^{2}t+bt}
$$
The coefficient function $K$ is determined by the initial condition
$$
                 f(x)=u(x,0)=\int_{-\infty}^{\infty}K(s)e^{isx}ds \\
                       K(s)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-isx}dx
$$
The final solution:
$$
   v(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\left(\int_{-\infty}^{\infty}f(y)e^{-isy}dy\right)e^{-s^{2}t+bt}\right]e^{isx}ds.
$$
The quantity in square brackets is $C=K(s)e^{-is^{2}t+bt}$, while the quantity in parentheses is $K(s)$. Now you use inverse Fourier transform and convolution results to simplify.
