Let's substitute $\alpha - 1$ with $\alpha - 1 = \beta$ to simplify the Fractional Differential Equation ($f^{\left( \alpha - 1 \right)} = f^{\left( \beta \right)}$):
$$
\begin{align*}
f^{\left( \alpha - 1 \right)}\left( t \right) &= t \cdot f\left( t \right)\\
f^{\left( \beta \right)}\left( t \right) &= t \cdot f\left( t \right)\\
\end{align*}
$$
Leonhard Euler came up with a nice method for some Fractional Derivatives (see Euler's Approach). He has sorted the derivatives of the function by order and is trying to find a pattern in it. He then tried to transfer this pattern from integer orders to fractional ones, which I always affectionately call the "Euler method". We can also use this procedure for some Fractional Differential Equation like this.
But for this we need a group of special functions, namely the Generalized Hypergeometric Functions $\operatorname{_{p}F_{q}\left( \cdots;\, \cdots;\, \cdot \right)}$ (I will abbreviate them to GHF in the following). The Generalized Hypergeometric Function has a definition given by:
$$
\begin{align*}
\operatorname{_{p}F_{q}\left( A_{1},\, A_{2},\, \cdots,\, A_{p};\, B_{1},\, B_{2},\, \cdots,\, B_{q};\, z \right)} &= \sum\limits_{k = 0}^{\infty}\left[ \frac{\left( A_{1} \right)_{k} \cdot \left( A_{2} \right)_{k} \cdots \left( A_{p} \right)_{k}}{\left( B_{1} \right)_{k} \cdot \left( B_{2} \right)_{k} \cdots \left( B_{q} \right)_{k} \cdot k!} \cdot z \right]\\
\end{align*}
$$
wehre $\left( x \right)_{y}$ is the Rising Factorial aka Pochhammer Symbol.
| $\beta$ |
Solution: $f(t)$ |
In the form of GHFs |
| $1$ |
$c \cdot e^{\frac{1}{2} \cdot t^{2}}$ |
$c \cdot \operatorname{_{0}F_{0}}\left[ \cdot;\, \cdot;\, \frac{1}{2} \cdot t^{2} \right]$ |
| $2$ |
$c_{1} \cdot \operatorname{Ai}\left( t \right) + c_{2} \cdot \operatorname{Bi}\left( t \right)$ |
$\cdots$ |
| $3$ |
$f\left( t \right)$ |
$c_{1} \cdot \operatorname{_{0}F_{2}}\left[ \cdot;\, \frac{1}{2},\, \frac{3}{4};\, \frac{1}{64} \cdot t^{\beta + 1} \right] + c_{2} \cdot t \cdot \operatorname{_{0}F_{2}}\left[ \cdot;\, \frac{3}{4},\, \frac{5}{4},\, \cdots;\, \frac{1}{64} \cdot t^{4} \right] + c_{3} \cdot t^{2} \cdot \operatorname{_{0}F_{2}}\left[ \cdot;\, \frac{5}{4},\, \frac{3}{2},\, \cdots;\, \frac{1}{64} \cdot t^{4} \right]$ |
| $4$ |
$f\left( t \right)$ |
$c_{1} \cdot \operatorname{_{0}F_{3}}\left[ \cdot;\, \frac{2}{5},\, \frac{3}{5},\, \frac{4}{5};\, \frac{1}{615} \cdot t^{5} \right] + c_{2} \cdot t \cdot \operatorname{_{0}F_{3}}\left[ \cdot;\, \frac{3}{5},\, \frac{4}{5},\, \frac{6}{5};\, \frac{1}{615} \cdot t^{5} \right] + c_{3} \cdot t^{2} \cdot \operatorname{_{0}F_{3}}\left[ \cdot;\, \frac{4}{5},\, \frac{6}{5},\, \frac{7}{5};\, \frac{1}{615} \cdot t^{5} \right] + c_{4} \cdot t^{3} \cdot \operatorname{_{0}F_{3}}\left[ \cdot;\, \frac{6}{5},\, \frac{7}{5},\, \frac{8}{5};\, \frac{1}{615} \cdot t^{5} \right]$ |
| $\vdots$ |
$\vdots$ |
$\vdots$ |
| $n$ |
$f\left( t \right)$ |
$\sum\limits_{k = 1}^{n \in \mathbb{N}}\left[ c_{k} \cdot t^{k - 1} \cdot \operatorname{_{0}F_{n - 1}}\left[ \cdot;\, \frac{k + 1}{n + 1},\, \frac{k + 2}{n + 1},\, \cdots;\, \frac{1}{\left( n + 1 \right)^{n}} \cdot t^{n + 1} \right] \right]$ |
where $\operatorname{Ai}\left( \cdot \right)$ and $\operatorname{Bi}\left( \cdot \right)$ are the $\operatorname{Ai}$- and $\operatorname{Bi}$- Airy Functions.
Note: The formulas marked in blue lead to Wolfram|Alpha, which confirms the results. Also Note: The series skips integer fractions.
We can generalize this simply by putting $\beta$ in a rounding function, such as the Ceil Function $\left\lceil \cdot \right\rceil$, Floor Function $\left\lfloor \cdot \right\rfloor$, or round $\operatorname{round}\left( \cdot \right)$.
$$
\begin{align*}
f\left( t \right) &= \sum\limits_{k = 1}^{\left\lfloor \beta \right\rfloor}\left[ c_{k} \cdot t^{k - 1} \cdot \operatorname{_{0}F_{\beta - 1}}\left[ \cdot;\, \frac{k + 1}{\beta + 1},\, \frac{k + 2}{\beta + 1},\, \cdots;\, \frac{1}{\left( \beta + 1 \right)^{\beta}} \cdot t^{\beta + 1} \right] \right]\\
\end{align*}
$$
So there is a solution given for $\alpha - 1 \in \mathbb{R}^{+}$ by:
$$
\begin{align*}
f\left( t \right) &= \sum\limits_{k = 1}^{\left\lfloor \alpha - 1 \right\rfloor}\left[ c_{k} \cdot t^{k - 1} \cdot \operatorname{_{0}F_{\alpha - 1 - 1}}\left[ \cdot;\, \frac{k + 1}{\alpha - 1 + 1},\, \frac{k + 2}{\alpha - 1 + 1},\, \cdots;\, \frac{1}{\left( \alpha - 1 + 1 \right)^{\alpha - 1}} \cdot t^{\alpha - 1 + 1} \right] \right]\\
f\left( t \right) &= \sum\limits_{k = 1}^{\left\lfloor \alpha - 1 \right\rfloor}\left[ c_{k} \cdot t^{k - 1} \cdot \operatorname{_{0}F_{\alpha - 2}}\left[ \cdot;\, \frac{k + 1}{\alpha},\, \frac{k + 2}{\alpha},\, \cdots;\, \frac{1}{\alpha^{\alpha - 1}} \cdot t^{\alpha} \right] \right]\\
\end{align*}
$$
So $\alpha - 1 \in \mathbb{R}^{+}$ gives:
$$\fbox{$
\begin{align*}
f\left( t \right) &= \sum\limits_{k = 1}^{\left\lfloor \alpha - 1 \right\rfloor}\left[ c_{k} \cdot t^{k - 1} \cdot \operatorname{_{0}F_{\alpha - 2}}\left[ \cdot;\, \frac{k + 1}{\alpha},\, \frac{k + 2}{\alpha},\, \cdots;\, \frac{1}{\alpha^{\alpha - 1}} \cdot t^{\alpha} \right] \right]\\
\end{align*}
$}$$
However, determining the whole thing for $\alpha - 1 \in \mathbb{R}^{-}$ is much more difficult but that's another topic. You can see with Wolfram|Alpha that the formula also works for higher derivatives like $\alpha = 21$.
