Solving the recurrence relation $x\to\frac{Kx}{x+(K-x)e^{-r}}+M$

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Suppose that $x_0$ is given and $$ x_n=\frac{Kx_{n-1}}{x_{n-1}+(K-x_{n-1})e^{-r}}+M,\qquad\text{ for }n\geq1 $$ where $K$, $r$, and $M$ are real constants. How can I find a general formula for $x_n$?

Context

Suppose that the initial population of a country is $x_0$, then the population of the country at time $t$, where $t$ is in years, is given by the logistic model

$$ x=\frac{Kx_0}{x_0+(K-x_0)e^{-rt}} $$ The logistic model does not include the immigration to the country. If we assume that on average $M$ people immigrate to the country each year, then how can we modify the logistic model so that it includes immigration to the country as well?

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The standard logistic growth model in discrete time is

$$N_{t+1}=N_t+rN_t\left(1-\frac{N_t}{K}\right)$$

where $N_t$ is the population at time $t$, $r$ is the natural growth rate, and $K$ is the carrying capacity.

To add immigration of amount $M$ each period just write:

$$N_{t+1}=N_t+rN_t\left(1-\frac{N_t}{K}\right)+M$$

However there is no closed form solution to this.


The standard logistic growth model in continuous time is

$$\frac{dN}{dt}=rN\left(1-\frac{N}{K}\right).$$

This has a nice closed form solution:

$$N(t)=\frac{K}{1+\left(\frac{K}{N_0}-1\right)e^{-rt}}$$

With constant immigration rate $M$ you would have

$$\frac{dN}{dt}=rN\left(1-\frac{N}{K}\right)+M$$

This has a closed form solution similar to the standard logistic growth model. You can find the solution in this paper here (as well as an extension to a stochastic model):

http://dx.doi.org/10.1016/j.tpb.2003.08.003