= Solution
This is a <birth-death process> with birth and death rates
$$
b_n=\gamma+\beta n,
\qquad
d_n=\alpha n(n-1).
$$
Its transition diagram has the two arrows
$$
n\xrightarrow{\ \gamma+\beta n\ }n+1,
\qquad
n\xrightarrow{\ \alpha n(n-1)\ }n-1.
$$
The gain into state $n$ comes from a death in state $n+1$ or a birth in state $n-1$, while the loss is the sum of both rates out of state $n$. The <birth-death master equation> is consequently
$$
\frac{\partial P(n,t)}{\partial t}
=\alpha n(n+1)P(n+1,t)
-[\alpha n(n-1)+\gamma+\beta n]P(n,t)
+[\gamma+\beta(n-1)]P(n-1,t).
$$
Multiplying by $n$, summing over the nonnegative integers, and shifting the summation indices shows that each birth contributes $+1$ and each death contributes $-1$. Thus the <first-moment equation of a birth-death process> gives
$$
\begin{aligned}
\frac{d\langle n\rangle}{dt}
&=\langle b_n-d_n\rangle\\
&=-\alpha\langle n^2\rangle
+(\alpha+\beta)\langle n\rangle+\gamma.
\end{aligned}
$$
At a stationary state, use the <variance> identity $\langle n^2\rangle=\langle n\rangle^2+(\Delta n)^2$. Solving the resulting <quadratic equation> gives
$$
\langle n\rangle
=\frac{\alpha+\beta}{2\alpha}
\mathbin{\pm}
\sqrt{\frac{(\alpha+\beta)^2}{4\alpha^2}
+\frac\gamma\alpha-(\Delta n)^2}.
$$
The minus root is inadmissible whenever it is negative, namely when $(\Delta n)^2<\gamma/\alpha$. Equality would give zero mean, which is also incompatible with a positive immigration rate $\gamma$, so for $\gamma>0$ the minus branch requires $(\Delta n)^2>\gamma/\alpha$ even to be a possible mean.
For a continuum approximation, write $b(n)=\gamma+\beta n$ and $d(n)=\alpha n(n-1)$. Applying the <Kramers-Moyal expansion> to the two gain terms and retaining derivatives through second order gives the <Fokker-Planck equation>
$$
\frac{\partial P}{\partial t}
=\frac{\partial}{\partial n}[g(n)P]
+\frac12\frac{\partial^2}{\partial n^2}[h(n)P],
$$
where the negative drift and infinitesimal jump variance are
$$
\begin{aligned}
g(n)&=d(n)-b(n)
=\alpha n^2-(\alpha+\beta)n-\gamma,\\
h(n)&=d(n)+b(n)
=\alpha n^2+(\beta-\alpha)n+\gamma.
\end{aligned}
$$
This truncation requires the typical population and the scale on which $P$ varies to be much larger than the unit jump size, with the rates varying smoothly across that scale.
The positive zero of $g$ is
$$
n_*
=\frac{\alpha+\beta+sqrt{(\alpha+\beta)^2+4\alpha\gamma}}{2\alpha}
=\frac\beta\alpha+1+\frac\gamma\beta+O(\alpha),
$$
so in particular $n_*\sim\beta/\alpha$ as $\alpha/\beta\to0$. Put $x=n-n_*$. The <linear noise approximation> uses
$$
g(n)=g_*'x+O(x^2),
\qquad h(n)=h_*+O(x),
$$
where
$$
g_*'=2\alpha n_*-(\alpha+\beta)
=\sqrt{(\alpha+\beta)^2+4\alpha\gamma}>0
$$
and, because $b(n_*)=d(n_*)$,
$$
h_*=h(n_*)=2b(n_*)=2(\gamma+\beta n_*).
$$
In a stationary state with zero <Fokker-Planck probability current>,
$$
g_*'xP+\frac{h_*}{2}\frac{dP}{dx}=0.
$$
Its normalized solution is the <normal distribution>
$$
P(x)\simeq
\sqrt{\frac{g_*'}{\pi h_*}}
\exp\left(-\frac{g_*'x^2}{h_*}\right).
$$
It follows that
$$
\langle n\rangle\simeq n_*\sim\frac\beta\alpha,
\qquad
(\Delta n)^2\simeq\frac{h_*}{2g_*'}
\sim\frac\beta\alpha.
$$
These estimates agree with the exact stationary first-moment relation on its plus branch to leading order: inserting $(\Delta n)^2\sim\beta/\alpha$ gives $\langle n\rangle\sim\beta/\alpha$. Moreover,
$$
\frac{\Delta n}{n_*}sim\sqrt{\frac\alpha\beta}\ll1,
\qquad
\Delta n\sim\sqrt{\frac\beta\alpha}\gg1.
$$
\b[Thus the stationary mass lies far from the boundary $n=0$, is narrow relative to its mean, yet changes across many lattice sites. These are precisely the large-population and slow-variation conditions needed for the diffusion approximation.]
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