Analytical solution of a single reservoir system with exponential outflow? The reservoir dynamic is the following:
$$
\frac{dS}{dt} = Q_{in} - Q_{out}(t)
$$
with
$$
Q_{out}(t) = k \times S(t)^\alpha
$$
where, 


*

*$S$ is the reservoir storage variable

*$t$ is time

*$Q_{in}$ is the constant inflow

*$Q_{out}(t)$ is the variable outflow

*$k$, $\alpha$ are positive parameters


Is it possible to find an analytical solution for S(t) with given S(t=0) ?
I don't know how to do this anymore without numerical solving. 

Edit:
Following @(Claude Leibovici)'s answer, analytical solution for $\alpha = 1$ and $\alpha = 2$ with respect to $S(t=0) = S_0$ are:


*

*$\alpha = 1$


$$
S(t) = \frac{Q_{in}}{k} + \big(S_0 - \frac{Q_{in}}{k}\big) e^{-kt}
$$


*

*$\alpha = 2$


$$
S(t) = \frac{\sqrt{Q_{in}}}{\sqrt{k}} tanh\big[\sqrt{Q_{in}}\sqrt{k}t+ tanh^{-1} \big(\frac{\sqrt{k}}{\sqrt{Q_{in}}} S_0 \big)\big)\big]
$$
 A: I do not think that you could get $S(t)$ except if $\alpha=1$ or $\alpha=2$.
However, writing $$\frac{dt}{dS}=\frac{1}{Q_{in}-k S^{\alpha }}\tag 1$$ the solution will be (unfortunately) 
$$t=C+\frac S {Q_{in}} \, _2F_1\left(1,\frac{1}{\alpha };\frac{1}{\alpha }+1;\frac{k S^{\alpha   }}{Q_{in}}\right)$$ where appears the hypergeometric function.
I am afraid that you will need numerical integration for the general case.
Edit
To show the steps, considering $(1)$, rewrite $$Q_{in}-k S^\alpha=Q_{in}\left(1-\frac k {Q_{in}} S^\alpha\right)$$ Now $$\frac k {Q_{in}} S^\alpha=U^\alpha \implies S=\left(\frac{Q_{in} }k\right)^{\frac 1 \alpha } U\implies dS=\left(\frac{Q_{in} }k\right)^{\frac 1 \alpha } dU$$ which make
$$dt=\frac1{Q_{in} }\left(\frac{Q_{in} }k\right)^{\frac 1 \alpha } \frac{dU}{1-U^\alpha}\implies  \frac{dU}{1-U^\alpha}=Q_{in}\left(\frac{Q_{in} }k\right)^{-\frac {1} \alpha }dt$$ $$\int \frac{dU}{1-U^\alpha}= Q_{in}\left(\frac{Q_{in} }k\right)^{-\frac {1} \alpha }t+C$$ Integrate the lhs to get $U$, go back to $S$ and you are done (for sure, adjust $C$ for the initial condition).
