I'm trying to solve the following problem:
Consider an M/M/$\infty$ queue in which the time intervals between task arrivals are i.i.d. exponential with parameter $\lambda$ and task durations are also i.i.d. exponential with parameter $\mu$. Show that, given that $n$ tasks are being processed at time $0$, the expected number of tasks being processed at time $t$ is $ne^{-\mu t} + \frac{\lambda}{\mu}(1 - e^{\mu t}).$
So far, I've found that, when the queue is in equilibrium, the average number of tasks being processed follows a Poisson distribution with parameter $\frac{\lambda}{\mu}$, but I can't make sense of how to arrive at this expression. How does one solve this?
Here is a partial solution:
Conditioned on there being $0$ tasks processed at time $0$, the differential-difference equations governing the system size are given by
\begin{align} p_n'(t) &= -(\lambda+n\mu)p_n(t)+\lambda p_{n-1}(t)+(n+1)\mu p_{n+1}(t),\ n>0\\ p_0'(t) &= -\lambda p_0(t)+\mu p_1(t). \end{align} The generating function of the probabilities $\{p_n(t)\}$ is given by $$ P(z,t) = \sum_{n=0}^\infty p_n(t)z^n = \exp\left((z-1)(1-e^{-\mu t})\frac\lambda\mu\right). $$ Expanding this as a Taylor series about $z=0$, we find that $$ p_n(t) = \frac1{n!}\left((1-e^{-\mu t})\frac\lambda\mu\right)\exp\left(-(1-e^{-\mu t})\frac\lambda\mu\right). $$ Computing the expectation we find $$ \sum_{n=0}^\infty n p_n(t) = \frac\lambda\mu(1-e^{-\mu t}). $$
From here it would remain to show that conditioned on there being $n$ tasks processed at time $0$, the expected number of tasks being processed at time $t$ is the above plus $n e^{-\mu t}$.