Does anyone have any clue of how to solve the following ODE:
\begin{align} \frac{F'(t)}{1-F(t)} &= p + (q-w)F(t) + w [F'(u) * e^{-\nu u}](t) \\ \frac{F'(t)}{1-F(t)} &= p + (q-w)F(t) + w \int_0^t F'(t-u) e^{-\nu u} du \\ \frac{F'(t)}{1-F(t)} &= p + (q-w)F(t) + w \int_0^t F'(u) e^{-\nu (t-u)} du \end{align}
After some algebra, I got the following equivalent ODE:
\begin{align} \frac{F'(t)}{1-F(t)} &= p + qF(t) - w\nu [F(u) * e^{-\nu u}](t) \\ \frac{F'(t)}{1-F(t)} &= p + qF(t) - w\nu \int_0^t F(t-u) e^{-\nu u} du \\ \frac{F'(t)}{1-F(t)} &= p + qF(t) - w\nu \int_0^t F(u) e^{-\nu (t-u)} du \end{align}
$F(0)=0$ and $*$ is the convolution operator.
I'm not sure whether it helps, but $F(t)$ is a cdf. So, $0 \le F(t) \le 1$, $lim_{t \rightarrow \infty} F(t) = 1$, and $F'(t) \ge 0$.
Thanks!
--
I tried using Laplace Transform in the second equation:
\begin{align} -\frac{d}{dt} \log(1-F(t)) &= p + qF(t) - w \nu [F(u) * e^{-\nu u}](t) \\ -\frac{d}{dt} \sum_{n=1}^{\infty} \frac{[F(t)]^n}{n} &= p + qF(t) - w \nu [F(u) * e^{-\nu u}](t) \end{align}
So:
\begin{gather} s \sum_{n=1}^{\infty} \frac{1}{n} \mathcal{L}\{F^n(t)\} = \frac{p}{s} + qY(s) - \frac{\nu w Y(s)}{s+\nu} \\ \mathcal{L}\{F(t)\} = Y(s) \end{gather}
But I'm stuck again.