This course page 12-13 (in French) is doing this :
For a low-pass filter, the equation is $$RCy't()+y(t)=x(t)$$ with distributions it's written $$ (-RC\delta'+\delta)*y=x $$ (I don't understand this well, even if I remember a bit the Dirac function and $<\delta, y'>=-<\frac {d\delta}{dx}, y>$ )
Then it is said the solution is $$y(t)=(h*x)(t) \text{, where } h(t)=\frac 1 {RC}e^{-t/RC} Y(t) \text{ (Y is the Heaviside step-function) } $$ (I don't understand how to reach this expression)
The distributional derivative of $h(t)$ is $$h'(t) = \frac{-1}{(RC)^2} e^{-t/RC}Y(t) + \frac{1}{RC}\delta_0 = \frac{-1}{RC}h(t) + \frac{1}{RC}\delta_0$$
You can check this by noticing that for any smooth function $\phi(t)$ with compact support, $$\int_{-\infty}^\infty h(t)\phi'(t)dt = - \int_{-\infty}^\infty \frac{-1}{(RC)^2}e^{-t/RC}Y(t)\phi(t)dt + \frac{1}{RC}\phi(0)$$ (To compute the integral on the left, note that you can change the bounds to $[0, \infty)$ for free, then integrate by parts. The boundary term at $\infty$ is zero since $\phi$ as compact support, and the boundary term at 0 is $\frac{1}{RC}\phi(0)$.)
Remember that the derivative of a convolution can be put on either product, so $$RC(h(t) \star x(t))' = RC\left(\frac{-1}{(RC)^2} e^{-t/RC}Y(t) + \frac{1}{RC}\delta_0\right) \star x(t))$$ so $$RC y'(t) = RC(h(t) \star x(t))' = -(h(t) \star x(t)) + x(t) = -y(t)+x(t)$$ and $$RC y'(t) + y(t) = -y(t)+x(t)+y(t) = x(t)$$ as desired.