General solution to a Growth equation I'd like to compute a formula that describes a population growth. The population starts with $N(t=0)$ individuals. At each time step there are births and deaths. The number of births at time $t$ is given by $N(t-1)f$, where $f$ being the fecundity. The number of deaths is given by $N(t-ls)$, where $ls$ being the lifespan. Therefore the recursive equation is:
$$
N(t)=N(t-1)+N(t-1) f-N(t-ls)
$$ 
which also equals:
$$
N(t)=N(t-1) (f+1)-N(t-ls)
$$
What is the general solution? I mean an equation that gives the value $N(t)$ in function  of $ls$, $f$ and $N(0)$ only.
Step-by-step solution is more than welcome.
UPDATE:
$ls$ and $t$ are integers.
 A: In principle, this is a simple constant-coefficient difference equation.  The problem is that $ls$ can be any positive integer, and this makes the equation difficult to solve specifically.  To wit, assume that the solution $N(t) = A r^t$, where $A$ is some constant and $r$ a growth rate.  Then we may find $r$:
$$r^t-(f+1) r^{t-1} +r^{t-ls}=0$$
or
$$r^{ls}-(f+1) r^{ls-1}+1=0$$
In general, this equation may be solved numerically for $r$, and produces $ls$ distinct values of $r$ i.e., $r_1, r_2, \ldots r_{ls}$.  Then the solution is
$$N(t) = \sum_{k=1}^{ls} A_k r_k^t$$
where the $A_k$ are found from, e.g., initial conditions.
A: Since your equation is a recurrence relation of order $ls$ (unless $ls=0$, in which case it's of order $0$, but I'll assume that that's not the case), the general solution is going to have $ls$ arbitrary constants (including $N_0$, but not including $f$). I kind of doubt you'll be able to get a "nice", explicit solution, but what I can get you is a "nice" ordinary generating function.
To make notation easier, I'll write $n$ instead of $t$, $N_n$ instead of $N(t)$, $r$ instead of $f+1$, and $a$ instead of $ls$. Then the recurrence relation is
$$N_n-rN_{n-1}-N_{n-a}=0.$$
Define $F(x):=\sum_{n=0}^\infty N_nx^n$ to be our ordinary generating function for $N_n$. Multiplying everything in the above equation by $x^n$ and summing from $n=a$ to $\infty$, we get
$$\begin{aligned}
0 &=\sum_{n=a}^\infty N_nx^n - r\sum_{n=a}^\infty N_{n-1}x^n + \sum_{n=a}^\infty N_{n-a}x^n\\
&= F(x)-(N_0+N_1x+\cdots + N_{a-1}x^{a-1})\\
&\quad{}-rx(F(x)-(N_0+N_1x+\cdots + N_{a-1}x^{a-2}))\\
&\quad{}+x^aF(x)\\
&= (1-rx+x^a)F(x) -(N_0+c_1x+c_2x^2+\cdots+c_{a-1}x^{a-1}),
\end{aligned}$$
where $c_1,\cdots,c_{a-1}$ are arbitrary constants. Then
$$ F(x)=\frac{N_0+c_1x+c_2x^2+\cdots+c_{a-1}x^{a-1}}{1-rx+x^a}.$$
Returning to the original notation, we get
$$ \sum_{t=0}^\infty N(t)\,x^t = \frac{N(0)+c_1x+c_2x^2+\cdots+c_{ls-1}x^{ls-1}}{1-(f+1)x+x^{ls}}.$$
A: Generating functions are your friend.
I will show how to get your generating function
from the recurrence.
I changed your notation a little.
You have
$n(t) = a n(t-1)-n(t-m)$,
or
$n(t+m) = a n(t+m-1)-n(t)$
.
Let
$f(x)
=\sum_{t=0}^\infty n(t) x^t
$.
$x^m f(x)
=\sum_{t=0}^\infty n(t) x^{t+m}
=\sum_{t=m}^\infty n(t-m) x^{t}
$
and
$x f(x)
=\sum_{t=0}^\infty n(t) x^{t+1}
=\sum_{t=1}^\infty n(t-1) x^{t}
$,
so
$\begin{align}
f(x)-ax f(x)+x^m f(x)
&=\sum_{t=0}^\infty n(t) x^t
-a\sum_{t=1}^\infty n(t-1) x^{t}
+\sum_{t=m}^\infty n(t-m) x^{t}\\
&=\sum_{t=0}^{m-1} n(t) x^t
-a\sum_{t=1}^{m-1} n(t-1) x^{t}
+\sum_{t=m}^\infty (n(t)-an(t-1)+n(t-m)) x^{t}\\
&=\sum_{t=0}^{m-1} n(t) x^t
-a\sum_{t=1}^{m-1} n(t-1) x^{t}\\
&= g(t)\\
\end{align}
$
where
$g(t)
=\sum_{nt0}^{m-1} n(t) x^t-a\sum_{t=1}^{m-1} n(t-1) x^{t}
=n(0)+\sum_{t=1}^{m-1} (n(t)-an(t-1)) x^t
$ 
is a polynomial of
degree $m-1$ that incorporates the
initial conditions of the $n(i)$.
Therefore
$f(x)
=\dfrac{g(x)}{h(x)}
$
where
$h(x) = 1-ax+x^m$.
The properties of the $n(i)$
thus depend on the initial conditions
(incorporated in $g(x)$)
and the roots of
$h(x)$.
In particular,
if $h(x)$ has no real roots,
the $n(i)$ will oscillate.
If $h(x)$ has positive real roots,
the $n(i)$
will grow
(unless there are special initial conditions)
like
$c/r^i$,
where $r$ is the root of $h$
of smallest value.
I will now try to determine what I can
about the roots of $h$.
I will assume that $a > 0$.
$h(0) = 1$.
Since $h'(x)
=-a+mx^{m-1}
$,
$h'(x)=0$
for
$x=x_0
=(a/m)^{1/(m-1)}
$,
so
$h'(x) < 0$
for
$0 \le x < x_0$
and
$h'(x) > 0$
for
$x > x_0$.
Therefore $h$ has a minimum at $x_0$.
To see if $h$ has any real root,
we need to see if
$h(x_0) < 0$.
If not,
$h$ has no real roots.
$\begin{align}
h(x_0)
&=1-ax_0+x_0^m\\
&=1-x_0(a-x_0^{m-1})\\
&=1-x_0(a-a/m)\\
&=1-(a/m)^{1/(m-1)}a(1-1/m)\\
&=1-(a/m)^{1+1/(m-1)}(m-1)\\
&=1-(a/m)^{m/(m-1)}(m-1)\\
\end{align}
$.
Therefore
$h(x_0) < 0$
if
$(a/m)^{m/(m-1)}(m-1) > 1$
or
$a > m/(m-1)^{(m-1)/m}$.
Therefore,
for any $m$,
if $a$ is large enough,
the $n(i)$
will grow exponentially.
Note that if $a = m/(m-1)^{(m-1)/m}$,
$h$ has a double root at $x_0$,
and the $n(i)$ will grow differently.
I'll take this a little further and leave it at that.
Let's see what happens for moderately large $m$.
$m/(m-1)^{(m-1)/m}
=\dfrac{m}{m-1}(m-1)^{1/m}
$.
$(m-1)^{1/m}
=e^{\ln(m-1)/m}
\approx 1+\ln(m-1)/m
$
and
$\dfrac{m}{m-1}
=\dfrac{1}{1-1/m}
\approx 1+1/m
$,
so
$m/(m-1)^{(m-1)/m}
\approx (1+\ln(m-1)/m)(1+1/m)
\approx 1+(1+\ln(m-1))/m
$.
Therefore
$a$ does not have to be
too large
for $h$ to have real roots.
That's enough for now.
