Forgive the terrible title, I can edit if someone has a better one! Also if this has been asked before, I wasn't sure exactly where to google or search for, so forgive me if this is a repeat thread. Likewise I didn't know what tags to use, so I'm happy to change them too.
Anyway, I'm stuck on how to solve the following equation (ie find a direct equation for $V(t)$ so that $V(t)$ does not appear on the right-hand side).
$$ \frac{dV}{dt} = \frac{1}{\alpha \ + \ \beta \cdot V(t) \ + \ \gamma \cdot V(t)\cdot V(t)},\ \ \ \text{s.t. } \ \ \ \alpha, \beta, \gamma > 0$$
$$V(0) = V_0, \ \ \ \text{s.t.} \ \ V_0> 0$$
Obviously we assume $\alpha, \beta, \gamma$ are positive constants, they don't depend on $t$.
(If anyone's curious, this equation will be used to feed into the following: $y = \frac{2V}{1+V} \in (0, 1)$. I'm trying to describe the organic growth rate of vegetation in a hexagonal tile in a simulated game.)
Does anyone have any tips on how to solve $V(t)$?
As an alternative question, which might be easier or harder (I don't know....) I would be happy with help in solving this variation instead:
$$ \frac{dV}{dt} = \frac{1}{\lambda + \texttt{exp}(\mu V(t))},\ \ \ \text{s.t. } \ \ \mu > 0, \ \lambda > -1$$
$$V(0) = V_0, \ \ \ \text{s.t.} \ \ V_0> 0$$
Your differential equation is separable, so you can write
\begin{align} & \frac{dV}{dt} = \frac{1}{\alpha + \beta V + \gamma V^2} \\ \implies & (\alpha + \beta V + \gamma V^2)dV = dt \end{align}
and integrate both sides:
\begin{align} \implies & \int (\alpha + \beta V + \gamma V^2)dV = \int dt \\ \implies & \alpha V + \frac 12 \beta V^2 + \frac 13 \gamma V^3 = t + C \\ \end{align}
where $C$ is a constant. Using the initial condition $V(0) = V_0$, we find that $C = \alpha V_0 + \frac 12 \beta V_0^2 + \frac 13 \gamma V_0^3$. Hence
$$\alpha V + \frac 12 \beta V^2 + \frac 13 \gamma V^3 = t + \alpha V_0 + \frac 12 \beta V_0^2 + \frac 13 \gamma V_0^3$$
Of course, you can use something like Cardano's substitution to solve explicitly for $V(t)$, but that will be a real mess.
Your alternative equation can be solved using the same method, but I don't think it gives an explicit expression for $V(t)$ either.