Well, I've to solve the following DE:
$$y(t)=x(t)\cdot\text{a}+\text{b}\cdot\ln\left(1+\frac{x(t)}{\text{c}}\right)+\int_0^tx(\tau)\cdot p(t-\tau)\space\text{d}\tau\tag1$$
And I've no idea how to start.
Some background information: this DE describes a current in an electric circuit. For the values of $a,b$ and $c$ I know that they are real and positive. For $c$ I know that $10^{-16}\le c\le10^{-4}$. The function $y(t)=k\cdot\theta\left(t-m\right)+(n-k)\cdot\theta\left(t-v\right)$ where all the values of the constants are ral and positive and $\theta$ is the Heaviside Theta function. The function $p(t-\tau)=\mathcal{L}_\text{s}^{-1}\left\{\frac{1}{\frac{1}{R_2+sl}+\frac{s}{sR_3+\frac{1}{z}}}\right\}_{\left(t-\tau\right)}$ where all the constants are again real and positive.
I almost certainly sure, that this integral equation has no analytic solution, since log makes it nonlinear. However, numerically speaking it's not that hard to solve.
First, we need to reverse non-linear part of the equation $G(x) = ax+b\ln(1+x/c)$. Since both terms are monotonously increasing, there's a unique inverse $G^{-1}(x)$.
So no we can, for example, find $x(0) = G^{-1}(y(0))$. It is an important step to use in solving the integral equation as Cauchy equation.
We take derivative from both sides. I will use $y'(t)=0$, since it's true almost everywhere, except for jumps in $t=m,v$, which I discuss later. $$ 0=ax'(t)+\frac{bx'(t)}{c+x(t)}+x(t)p(0)+\int_0^tx(\tau)p'(t-\tau)d\tau,\\ x'(t)=-\left(a+\frac{b}{c+x(t)}\right)^{-1}\left(x(t)p(0)+\int_0^tx(\tau)p'(t-\tau)d\tau\right). $$
We expressed the derivative $x'(t)$ as a function of $t,x(t)$, so we can use any method for solving ODE (it would be longer than usual, since integral, but still very doable on modern machines). You can precalculate $$p'(t-\tau)=\mathcal L^{-1}_s\left\{\mathcal L_s p'(t-\tau) \right\}_{t-\tau}= \mathcal L^{-1}_s\left\{\mathcal s P(s)-p(0^{+}) \right\}_{t-\tau}=\\ \mathcal L^{-1}_s\left\{\frac{s}{\frac{1}{R_2+\ldots}+\ldots} \right\}_{t-\tau} - p(0^+)\delta(t-\tau)$$
and since $0<\tau<t$, we can neglect Dirac delta.
The only thing left to discuss is what to do with jumps in $y(t)$. I would assume, $m<v$. Say we have found the solution for $x(t)$ from $0$ to $m^{-}$ (right before the jump). Then we can calculate the value $x(m^+)$ (after the jump), using the same technique, we used to calculate $x(0)$:
$$ y(m^+) = G(x(m^+)) + \int_0^m x(\tau)p(m-\tau)d\tau,\\ x(m^+) = G^{-1}\left( y(m^+) - \int_0^m x(\tau)p(m-\tau)d\tau \right) $$
So it goes like this: you calculate $x_0=x(0)$, then solve ODE from $0$ to $m$, then calculate value $x_1=x(m^+)$, then solve ODE from $m$ to $v$, then calculate value $x_2=x(v+)$, and finally solve ODE from $v$ to whatever you need.
Edit. I just noticed that your function $p(t-\tau)$ has Dirac delta component $R_3\delta(t-\tau)$, which lies on the end of integration interval. Since you use Laplace transform, I believe the convolution arises from the similar matters. In that case, I believe $\int_{0}^t$ is really $\int_{0^-}^{t^+}$, so domain of delta function lies inside the integration domain. Thus, you need to add manually $x(t)R_3$ to integrals.