in the following Paper a ODE model of a zombie infection is given. \begin{equation} \begin{aligned} \frac{dS}{d\tau} &= -\frac{SZ}{N} \\ \frac{dZ}{d\tau} &= (1-\alpha)\frac{SZ}{N}\\ \frac{dR}{d\tau} &= \alpha\frac{SZ}{N} \\\\ N &= S + Z + R = const \quad \dots\\ \text{initial condition: } S(0) &= S_0,\, Z(0) = Z_0,\, R(0) = 0 \end{aligned} \end{equation} The system is non-linear and there exits a analytical solution.
\begin{equation*} \begin{aligned} P &\equiv Z_0 + (1-\alpha)S_0 \\ \mu &\equiv \frac{P}{Z_0}-1\\ f(\tau) &\equiv \frac{P\mu}{\exp{(\tau P/N)} + \mu}\\ Z(\tau) &= P - f(\tau) \\ S(\tau) &= \frac{f(\tau)}{1-\alpha} \end{aligned} \end{equation*}
Unfortunatly I don't know how to solve the system to get the analytical solution. My Idea is to substitute X = SZ but if I calculate $\dot{X}$ I can't eliminate S or Z from the equation. \begin{equation} \dot{X} = \dot{S}Z + \dot{Z}S \end{equation} I'm out of ideas how to solve the system and I hope for some suggestions
The clue is in the definitions you give. Define $P(\tau)=Z(\tau)+(1-\alpha)S(\tau)$, then
$$\frac{dP}{d\tau} \,=\, \frac{dZ}{d\tau} + (1-\alpha)\frac{dS}{d\tau} \,=\, (1-\alpha)\frac{SZ}{N} - (1-\alpha)\frac{SZ}{N} \,=\, 0.$$
Thus $P(\tau)$ is constant so from the initial conditions: $P(\tau)=Z_{0}+(1-\alpha)S_{0}$. Now from our definition of $P(\tau)$:
$$ (1-\alpha)S \,=\, P-Z \quad\Longrightarrow\quad (1-\alpha)\frac{SZ}{N} \,=\, \frac{(P-Z)Z}{N} \quad\Longrightarrow\quad \frac{dZ}{d\tau} \,=\, \frac{(P-Z)Z}{N}. $$
This is now an ODE for $Z(\tau)$ since $P,N$ are constant. I presume you can solve this subject to the initial conditions; it has solution as given above where the function $f(\tau)$ and constant $\mu$ are introduced so that $Z(\tau)=P(\tau)-f(\tau)$ - a much simplified expression. In particular, note that this implies $P(\tau)-Z(\tau)=f(\tau)$. Then the final solution for $S(\tau)$ follows again from the definition of $P(\tau)$:
$$(1-\alpha)S\,=\,P-Z \quad\Longrightarrow\quad S\,=\,\frac{P-Z}{1-\alpha} \,=\, \frac{f(\tau)}{1-\alpha}.$$
I notice no solution for $R(\tau)$ is given; presumably these can be found since we now have explicit expression for $R(\tau)$.