Let us denote the steady-state number of infectives by $\hat{I}$ and the steadystate number of susceptibles by $\hat{S}$, where $\hat{I}$ and $\hat{S}$ are constants. Then,
$$
I_{k+1}=I_{k}=\hat{I} \text { and } S_{k+1}=S_{k}=\hat{S}
$$
are the steady-state solutions. Substituting (4) into (3) we obtain
$$
\hat{I}=f \hat{S} \hat{I}
$$
$$
\hat{S}=\hat{S}-f \hat{S} \hat{I}+B,
$$
or
$$
\begin{aligned}
\hat{I}(1-f \hat{S}) &=0 \
f \hat{S} \hat{I}-B &=0 .
\end{aligned}
$$
Our aim is to solve for $\hat{I}$ and $\hat{S}$. From equation (5a) there are two cases to be considered: $\hat{I}=0$ or $\hat{S}=1 / f$.
MTH2032 COURSE NOTES ：
$$
Y_{k+1}=(1-r) Y_{k}
$$
The solutions of the linear differenceare only approximate solutions of but they have the advantage that they can be found in closed form.
The closed-form solution of is
$$
Y_{k}=(1-r)^{k} Y_{0}
$$
where $Y_{0}$ can be calculated from the initial population $N_{0}$ using. Hence an approximate solution to the original is given by
$$
N_{k} \simeq K+K(1-r)^{k} Y_{0}
$$