Poisson Process Derivation. I was looking at a derivation for the poisson process , which tells the number of events occurring in time $t$ , 
I came across the following differential equation :
$\frac{d}{dt}(P_n(t))$ = $\lambda[P_{n-1}(t) - P_n(t)]$ , but i have no idea how to tackle this differential equation , 
Here , $P_n(t)$ = Probability that $n$ events occur in time interval $t$.
Probability that $0$ events occur in time interval $t$ is given as : $P_0(t)$ = $e^{-\lambda t}$ ,
Also , how can we take $P_n(t+dt)$ = $P_n(t)$ + $dt*\frac{d}{dt}(P_n(t))$ , where ,
$P_n(t+dt)$ is the probability that $n$ events occur in the time interval $t+dt$.
Kindly help me with this...
 A: So, let's cope first with this differential equation:
$$ \frac{d}{dt} P_n(t) = \lambda \left (P_{n-1}(t) - P_n(t) \right)$$
or
$$ \frac{d}{dt} P_n(t) = - \lambda P_{n}(t) + \lambda P_{n-1}(t).$$
We can solve these equations step by step: $P_0 (t)$ is known, then we plug it in equation for $P_1(t)$ and find it, and so on.
The ODE that we have here is first-order linear inhomogeneous equation. It has the form
$$\frac{dx}{dt} = \alpha x + g(t)$$
and solution to the Cauchy problem $x(0) = x_0$ is 
$$ x(t) = x_0 e^{\alpha t} + e^{\alpha t} \cdot \int\limits_{0}^{t} e^{-\alpha \tau} g(\tau) \, d\tau . $$
This solution can be found either from multiplying by integration factor $e^{-\alpha t}$ or by treating this equation as linear (find solution of homogeneous equation and do the variation of parameters -- this would give you a general solution).
Let's apply this knowledge to your particular ODE. Put $\alpha = - 
\lambda$; then the solution can be written this way:
$$ P_n(t) = P_n(0) e^{-\lambda t} + e^{-\lambda t} \cdot \int\limits_{0}^{t} e^{\lambda \tau} \cdot \left( \lambda P_{n-1}(\tau) \right ) \, d\tau$$
As André Nicolas mentioned, the rest can be done by induction. What I've obtained is that 
$$P_n(t) = e^{-\lambda t} \cdot \sum\limits_{k=0}^{n} \left ( P_k(0)\cdot\frac{(\lambda t)^{n-k}}{(n-k)!} \right ), $$
hope that this is close to truth :)
