$\mathbf{Task}$: M/M/1/1 queue where user who get serviced reenters the queue with probability $\alpha$(going to the beginning of the queue). Find limiting probabilities.
$\mathbf{My}$ $\mathbf{attempt}$:
Differential equations: $$\begin{cases} \frac{dP_0(t)}{dt} = - \lambda_eP_0 + \mu P_1 \\ \frac{dP_1(t)}{dt} = \lambda_eP_0 - \mu P_1 \end{cases} $$ where $\lambda_e = \lambda + \lambda (1 - P_{loss}) \alpha$. So, as we need limiting probabilities $\lim_{t\to\infty}{\frac{dP_i(t)}{dt}} = 0$ and equations will be like this $$\begin{cases} 0 = - \lambda_ep_0 + \mu p_1 \\ 0 = \lambda_ep_0 - \mu p_1 \end{cases} $$ Then $P_{loss} = P_1$ because it describes part of time when all services are "busy". So $p_{loss} = p_1$ in the limit. And now our equations: $$\begin{cases} 0 = - (\lambda + \lambda (1 - p_1) \alpha)*p_0 + \mu p_1 \\ 0 = (\lambda + \lambda (1 - p_1) \alpha)*p_0 - \mu p_1 \end{cases} $$ Solving this system we get $p_1 = \frac {p_0 (\lambda + \lambda \alpha)}{\lambda\alpha p_0 + \mu}$ . Then using the fact $p_0 + p_1 = 1$ we can solve quadratic equation and get something like this $p_0 = \frac{-(\mu + \alpha)^2 + \sqrt{\mu^2 +2\alpha\mu + \alpha^2 + 4\mu-4\lambda\alpha}}{2\lambda\alpha}$
$\mathbf{Question}$: Is this solution correct? If you look at the $p_0$, it seems like the solution is wrong :)
Since a customer reenters with probability $\alpha$ and each time has service time exponentially distributed with mean $\frac1\mu$, the sojourn time of a customer is exponentially distributed with mean $\frac\mu\alpha$ (deriving this is a straightforward computation).
So we have a continuous-time Markov chain with generator matrix $$ \left( \begin{array}{cc} -\lambda & \lambda \\ \alpha \mu & -\alpha \mu \\ \end{array} \right), $$ from which it is straightforward to compute the unique stationary distribution $$ \left(\frac{\alpha\mu}{\lambda+\alpha\mu},\frac{\lambda}{\lambda+\alpha\mu}\right). $$