Consider the homogeneous Volterra equation
$f(x) = \int_0^x dx' K(x,x')f(x')$
and assume that the kernel is separable, $K(x,x')=m(x)n(x')$ leading to
$f(x) = m(x)\int_0^x dx' n(x')f(x')$.
Assume that we know the function $G(x)$ so that $G'(x) = n(x)f(x)$. We can write
$f(x) = m(x) [G(x)-G(0)]$.
Next, I differentiate both terms with respect to x, obtaining
$f'(x) = m'(x) [G(x)-G(0)]+m(x)n(x)f(x)$,
and I replace $[G(x)-G(0)]$ by $f(x)/m(x)$. This finally leads to
$f'(x) = [\frac{m'(x)}{m(x)}+m(x)n(x)]f(x)$.
So, apparently, I can solve my original problem by solving this differential equation. I now try a specific form of the kernel, $K(x,x') = e^{A(x-x')}$. By applying the last formula, I get the differential equation
$f'(x) = [A+1]f(x)$
which has the solution $f(x)=e^{(A+1)x}f(0)$. I now plug this solution into my original problem,
$f(x) = e^{Ax}\int_0^x dx' e^{-Ax'}e^{(A+1)x'}f(0)=e^{Ax}f(0)\int_0^x dx'e^{x'} = e^{(A+1)x}f(0) - e^{Ax} f(0)$,
which doesn't match.
I assume the problem originates somewhat from the fact that I differentiated the original problem, thus possibly adding "fake" solutions. Can you help me understanding it better? The solutions would match if $f(0)=0$, leading to $f(x)=0$. Does this mean that the original problem admits only the solution $f(x)=0$?