Fractional Bernoulli equation and logistic function I'm investigating the solution of the special case of the Bernoulli differential equation
$$
\frac{dy}{dt} = \frac{y(1-y)}{\tau}, \tag{1}
$$
with $\tau$ a time constant, and which models innovation processes fairly well, and whose solution is the logistic curve
$$ y = \frac{1}{1+e^{-t/\tau}}.\tag{2}$$
Recently I've become acquainted with fractional calculus, and I'm interested in finding out if the solution to the corresponding fractional Bernoulli differential equation
$$D^{\alpha} y = y(1-y)/\tau, \ \alpha\in(0,1) \tag{3}$$
is given by substituting the exponential with the one-parameter Mittag-Leffler function
$$E_{\alpha}(-t^\alpha/\tau) \tag{4}$$
that is
$$ y = \frac{1}{1+E_{\alpha}(-t^\alpha/\tau)}. \tag{5}$$
I've consulted a couple of papers that for ordinary differential equations show an almost direct correspondance between exponential and Mittag-Leffler, but do not have a solution to my question.
I would be grateful for pointing me to the correct solution and a paper showing it.
Mainardi, Francesco, On some properties of the Mittag-Leffler function (E_\alpha(-t^\alpha)), completely monotone for (t>0) with (0<\alpha<1), Discrete Contin. Dyn. Syst., Ser. B 19, No. 7, 2267-2278 (2014). ZBL1303.26007.
Luchko, Yuri, [Operational method for fractional ordinary differential equations; in Handbook of fractional calculus with applications], vol.2 - Fractional Differential Equations](2022).
 A: --- EDIT ---
I was told by one of the authors of the papers I cited that the chain rule (6) on which I based my proof is wrong. See this article. There are a number of articles in fractional calculus that have publicized formulas that were in error; but unfortunately have been accepted by the peer review (reviewers either do not sufficiently understand the field, or do not check the articles). Worse, the articles have not been retracted. Bottom line my writing below is not a proof. The text below stands for now, as a reminder
of the difficulty, and waiting for an appropriate correction.
--- END of EDIT ---
Being a beginner in Fractional Calculus, I had to read more than a couple of papers and watch a few videos to understand the methods and arrive at an answer. The tentative solution proposed in the OP is correct. The proof for $\tau=1$ consists in calculating both sides of the equation separately, and then checking if they coincide.
For the left-hand side of equation (1), we can calculate the Caputo Fractional derivative of equation (5), by using the chain rule, which in Fractional Calculus is
$${}^C D^\alpha f[g(x)]\,=\,\frac{df}{dg}\cdot D^\alpha g(x) \tag{6}$$
Thus adopting the notation by Mainardi in his paper,
$$ E_\alpha (t^\alpha)\,=\,e_\alpha (t), \tag{7}$$
and considering that in our case expression (5) can be thought as composed of the following
$$\begin{align}f &=  \frac 1 {1+g} \tag{8} \\
g &= e_\alpha (-t), \tag{9}
\end{align}$$
then
$$\begin{align}\frac{df}{dg} &=  -\frac 1 {{(1+g)}^2} \tag{10} \\
{}^C D^\alpha g(x) &= {}^C D^\alpha e_\alpha (-t) \, = -e_\alpha (-t), \tag{11}
\end{align}$$
from which
$$\begin{align} {}^C D^{\alpha} y &= -\frac 1 {{(1+g)}^2} \cdot {}^C D^\alpha g(x) \; && \text{substituting (9) and (11)}\\
&= -\frac 1 {{(1+e_\alpha (-t))}^2} \cdot \left( -e_\alpha (-t) \right) 
&&\text{simplified}\\
&= \frac {e_\alpha (-t)}{{[1+e_\alpha (-t)]}^2} \tag{12}
\end{align}$$
Similarly for the right-hand side of equation (1), we can calculate
$$\begin{align}
y(1-y)/\tau &= &&\text{using (8)}\\
&= \frac 1 {1+g} \left(1-\frac 1 {1+g}\right) &&\text{simplifying}\\
&= \frac g {{(1+g)}^2} &&\text{substituting (9)}\\
&= \frac {e_\alpha (-t)}{{[1+e_\alpha (-t)]}^2} \tag{13}
\end{align}$$
