Question: How could we solve this ODE system $$u'=\alpha(1-x)-(\alpha+\beta)u, \quad u(0)=u_0 $$ $$v'=\gamma x-(\gamma+\delta)v,\quad v(0)=v_0 $$ $$x'=\delta v(1-x)-\beta ux, \quad x(0)=x_0 $$ Here, $\alpha,\beta,\gamma,\delta>0$ and we can assume that $\alpha\ll \beta, \gamma\ll \delta$?
My attempt:The equilibria are $(\frac{\alpha}{\alpha+\beta},0,0)$ and $(0,\frac{\gamma}{\gamma+\delta},1)$. I would like to find explicit solutions or good approximations. Numerically, $u(t),v(t),x(t)$ look like logistic functions.
The best I came up with, is the approximation $$u\approx\frac{\alpha}{\alpha+\beta}(1-x),\quad v\approx\frac{\gamma}{\gamma+\delta} x$$ when $u$ and $v$ are small compared to $x$ and hence their derivatives $\dot{u}\approx 0,\dot{v}\approx 0$. This implies $$\dot{x}=\underbrace{\left(\frac{\gamma\delta}{\gamma+\delta}-\frac{\alpha\beta}{\alpha+\beta}\right)}_{\equiv c}x(1-x)$$ so $$x(t)=\frac{x_0e^{ct}}{1+x_0(e^{ct}-1)},\quad u(t)= \frac{\alpha}{\alpha+\beta}(1-x(t)),\quad v(t)=\frac{\gamma}{\gamma+\delta} x(t)$$ Problem: While the limiting behavior of the approximation is the same as the exact solution, at intermediate times there is a substantial discrepancy. So I'm seeking better approximations or even an exact solution.
Any help or comment is greatly appreciated!