Consider the PDE
$$ \partial\_t f(t,x) = \Phi(x) f(t,x)+ \Psi(x)f(t,x-a)$$
$$f(0,x)=1$$
$$f(t,0)=1$$
where $a$ is a constant and $\Phi$ and $\Psi$ are some differentiable functions. To solve this I used separation of variables $f(t,x)=g(x)h(t)$ and I get
$$ \frac{\partial\_t h(t)} {h(t)}=\Phi(x)+\Psi(x)\frac{g(x-a)}{g(x)}=c\_i$$ 


Where $c\_i$ is a constant so we get $h(t)=d\_i e^{c\_i t}$ and we obtain 
$$ g(x)=\prod\_{k=0}^{\infty} \frac{\Psi(x-ka)}{c\_i-\Phi(x-ka)}$$ 
so we have two free parameter $c\_i$ and $d\_i$ so the question is what is the general solution of the PDE based on these two free parameters? Is that correct to write the general solution as 
$$ \int\int z e^{yt} \prod\_{k=0}^{\infty} \frac{\Psi(x-ka)}{y-\Phi(x-ka)} dy dz\quad ? $$ 



Separation of variables for PDE
Assuming everything converges I think the correct candidate for a general solution for your PDE (leaving boundary conditions aside) is rather
$$
\int q(y) e^{yt} \prod\_{k=0}^{\infty} \frac{\Psi(x-ka)}{y-\Phi(x-ka)} dy 
$$
where $q(y)$ is an arbitrary function of $y$.


Formally, it satisfies your PDE (again assuming everything converges and you can differentiate under the integral and product signs) and contains an arbitrary function of one variable, and so should be a general solution.


