$\frac{\partial v(x,t)}{\partial t}- C\frac{\partial^2 v(x,t)}{\partial x^2}=b{v(x,t)}$
$IC: v(x,0) = f(x)$
$ -\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.
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.