The integral equation for (causal) convolution is given by
$$y(t) = \int_{-\infty}^{t} K(t - \tau) x(\tau) d\tau$$
Can one write an equivalent differential equation for general well-behaved kernel $K$?
Since convolution with an exponential kernel is a solution to a linear ODE with with time-dependent input (e.g. see 2nd example here), the transformation is possible for some kernels. If it turns out that there are well-behaved kernels for which the transformation is not possible, it would be great to address the question on determining the set of kernels for which the transformation is possible.
EDIT: I have found a related post, where respondents argue that this operation is indeed impossible for general kernels. I will comment further when I have fully understood if the outlined arguments apply to my case, where causality is enforced by the limits of integration.
Let me elaborate on my comment:
rewrite $y(t) = \int_{-\infty}^{t} K(t - \tau) x(\tau) d\tau$ as
$y(t) = \int_{-\infty}^{\infty} \Theta(t-\tau)K(t - \tau) x(\tau) d\tau$ for $\Theta(x)$ the Heavyside function.
Call $\Theta(x) K(x) = R(x)$.
Then a Fourier transform yields:
$\hat y(\omega) = \hat R(\omega) \hat x(\omega) $ . If $\hat R(\omega) = {P(\omega)\over Q(\omega)} $ then
$Q(\omega) \hat y(\omega) = R(\omega) \hat x(\omega)$
In the inverse Fourier we may interpret $i\omega\rightarrow {d\over dt}$ to get $Q( {d\over dt}) y(t) = R( {d\over dt})x(t) $ , a differential equation.
For example :
$y(t) = \int_{-\infty}^{t} e^{-k(t - \tau)} x(\tau) d\tau$ yields
$\hat y(\omega) = {1\over i\omega +k} \hat x(\omega)$
$\rightarrow$
$({d \over dt} +k)y(t)=x(t)$
I am not claiming that only kernels whose transforms are polynominals of $\omega$ can be transformed into differential equations, but it is quite a broad family.