general solution to fractional differential equation I'd like to know the existence of the general solution to the following fractional differential equation $$D_{0+}^{\alpha} y(t)=0  \text{,}\label{1} \tag{1}$$
where $\alpha \in (1,2)$ and
$$  D_{0+}^{\alpha} y(t)=\frac{1}{\Gamma(2-\alpha)} \left( \frac{d}{dt} \right)^2 \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds  \text{.}  $$
I know that $t^{\alpha-1}$ and $t^{\alpha-2}$ are solutions to problem \eqref{1}.
My question is whether $C_1t^{\alpha-1} + C_2 t^{\alpha-2}$ is the general solution to problem \eqref{1}.
Note that $D_{0+}^{\alpha}$ has second order derivative.
My attempt:
Let $Y(t)=\int_0^t \frac{y(s)}{(t-s)^{\alpha-1}}$ for fixed $y$. Then
$y$ is a solution to problem \eqref{1} if and only if $Y$ is a solution to problem $Y''(t)=0$.
We also know that

*

*$D_1+D_2 t$ : a general solution to problem $Y''(t)=0$.


*$y(t)=t^{\alpha-1} \implies Y(t)=Ct$ and  $y(t)=t^{\alpha-2} \implies Y(t)=C$ for a constant $C=\int_0^1 (1-v)^{1-\alpha}dv. $
I think I can prove it using them, but I'm not sure.
I would be grateful if you could give any comment for my question.
 A: Suppose $y: \mathbf R^+\rightarrow \mathbf R$ is such that $\int_0^\infty |f(t)e^{-at}|dt<\infty$ for some $a\ge0$. You can apply the Laplace transform $\mathcal L$ on $Y$ which is defined by OP in his question since
$$Y(t) = y*t^{1-\alpha}$$
where $*$ stands for the convolution operator, and $Y$ is a power function thus converges under the Laplace transform. We will repeatedly use the property that
$$\mathcal L\{t^q\}(s)=\frac{\Gamma(q+1)}{s^{q+1}},\quad \forall t>0,\,\text{Re}(s)>0,\, \text{Re}(q)>-1.$$
and the transformation is invertible. Apply the Laplace transform on $Y=D_1+D_2t$.
$$\mathcal L\{Y\} = \mathcal L\{y\}\mathcal L\{t^{1-\alpha}\}=\mathcal L\{y\}\frac{\Gamma(2-\alpha)}{s^{2-\alpha}}=\frac{D_1}{s}+\frac{D_2}{s^2},$$
then
$$\mathcal L\{y\}(s)=\frac{a_1}{s^{\alpha-1}}+\frac{a_2}{s^\alpha}.$$
Take the inverse Laplace transform
$$y(t)=b_1t^{\alpha-2}+b_2t^{\alpha-1}.$$
$a$'s and $b$'s above are constants.
A: $\color{green}{\textbf{Version of 30.10.21.}}$
Starting from the Riemann-Liouville fractional integral in the form of
$$\text D_{0+}^{-\beta} f(t)=\dfrac1{\Gamma(\beta)} \int\limits_0^t (t-x)^{\beta-1} f(x)\,\text dx,\quad\text{where}\quad \text D=\dfrac{\text d}{\text dt},\quad \beta\in(0,1)\tag1$$
(see the link from the OP comments), and assuming $\,\beta=2-\alpha,\,$ one can get
$$\text D_{0+}^{\alpha-2}y(t)=\dfrac1{\Gamma(2-\alpha)}\int\limits_0^t (t-x)^{1-\alpha}y(x)\,\text dx, \quad\text{where}\quad \alpha\in(1,2),\tag2$$
wherein
$$\text D_{0+}^{\alpha-2}t^{\omega}
=\dfrac1{\Gamma(2-\alpha)}\int\limits_0^t x^{\omega}(t-x)^{1-\alpha}\,\text dx
=\dfrac{t^{\omega-\alpha+2}}{\Gamma(2-\alpha)}
\int\limits_0^t \left(\dfrac xt\right)^{1-\alpha}\left(1-\left(\dfrac xt\right)\right)^{\omega}\,\text d\left(\dfrac xt\right),$$
$$\text D_{0+}^{\alpha-2}t^{\omega}
=\dfrac{\operatorname B(2-\alpha,\omega+1)} {\Gamma(2-\alpha)} \,t^{\omega-\alpha+2}
=\dfrac{\Gamma(\omega+1)} {\Gamma(\omega-\alpha+3)} \,t^{\omega-\alpha+2},\tag3$$
$$\text D_{0+}^{\alpha-1}t^{\omega}
=\begin{cases}
0,\quad\text{if}\quad \omega=\alpha-2\\[4pt]
\dfrac{\Gamma(\omega+1)}{\Gamma(\omega-\alpha+2)} \,t^{\omega-\alpha+1}, \quad\text{otherwize},
\end{cases}\tag4$$
$$\text D_{0+}^{\alpha}t^{\omega}
=\begin{cases}
0,\quad\text{if}\quad \omega\in\{\alpha-2,\alpha-1\}\\[4pt]
\dfrac{\Gamma(\omega+1)}{\Gamma(\omega-\alpha+1)} \,t^{\omega-\alpha},
\quad\text{otherwize}.
\end{cases}\tag5$$
Used approach and formulas $(3)-(5)$ allow to solve certain linear ODEs, which contain fractional derivatives.
In particular, the given ODE has the common solution in the form of
$$y(t)=C_1e^{(\alpha-1)t}+C_2e^{(\alpha-2)t}.$$
A: Only integration, differentiation, and algebra answer:
Note that you use an unfamiliar definition of the fractional derivative. Let me try using the inverse of the fractional derivative called which can be represented via Fractional Integral and Cauchy’s formula for repeated integration:
Your equation is simply:
$$\frac{1}{\Gamma(2-\alpha)} \left( \frac{d}{dt} \right)^2 \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds=0 $$
Let’s integrate twice wrt $t$ and ignore all other constants:
$$\iint\frac{1}{\Gamma(2-\alpha)} \left( \frac{d}{dt} \right)^2 \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds (dt)^2=\iint\frac{1}{\Gamma(2-\alpha)} \left( \frac{d}{dt} \right)^2 \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds (dt)^2=\iint 0(dt)^2\implies \int\frac{1}{\Gamma(2-\alpha)} \left( \frac{d}{dt} \right)\int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds+c_0 dt=\int 0+c_1dt\implies \frac{1}{\Gamma(2-\alpha)} \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds+c_0t+c_3 = c_1t+c_2$$
Therefore we can combine constants:
$$ \frac{1}{\Gamma(2-\alpha)} \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds= c_1t+c_2$$
According to the Riemann–Liouville fractional integral section on Wikipedia, the nth integral is defined by:
$$\,_0D^{-\alpha}_t f(t) =\frac1{\Gamma(a)} \int_0^t(t-x)^{\alpha-1}f(x)dx$$
Therefore:
$$\,_0D^{-\alpha}_t  \frac{1}{\Gamma(2-\alpha)} \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds=\,_0D^{-\alpha}_t(c_0t+c_1)\implies \frac1{\Gamma(a)} \int_0^t(t-s)^{\alpha-1}\frac{1}{\Gamma(2-\alpha)} \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, dsds=\frac1{\Gamma(2-\alpha)\Gamma(a)} \int_0^t\int_0^t y(s) dsds=\frac1{\Gamma(a)} \int_0^t(t-s)^{\alpha-1}(c_0 s+c_1)ds$$
You can also just notice that the fractional derivative is just repeated differentiation, so we can just integrate both sides repeatedly and include the $\alpha$ as a constant:
$$ \int_0^t \frac{y(s)}{(t-s)^{\alpha-1}} \, ds= c_1t+c_2\implies \int_0^t  \int_0^t \frac{y(s)(t-s)^{\alpha-1}}{(t-s)^{\alpha-1}} \, dsds=\frac{\Gamma(\alpha)}{\Gamma(\alpha)}\int_0^t(c_0s+c_1)(t-s)^{\alpha-1}ds$$
Therefore:
$$\left(\int_0^t\right)^2y(s)(dt)^2=\int_0^t(c_0s+c_1)(t-s)^{\alpha-1}ds$$
If you take it like a repeated integral, you would get a solution as:$$y=c_0t^{\alpha-1}+c_1 t^\alpha=D^{-\alpha}_t (c_0 t+c_1)$$
Here is the actual solution after evaluation of the integral and differentiating twice:
$$\left(\int_0^t\right)^2y(s)(dt)^2=\int_0^t(c_0s+c_1)(t-s)^{\alpha-1}ds\implies y(s)=c_0t^{a-1}+(\alpha-1)c_1t^{\alpha-2}\implies \boxed{y(t)=c_0t^{\alpha-1}+c_1 t^{\alpha-2}} $$
Please correct me and give me feedback!
Let’s try something else and simplify your definition:
Note that:
$$\frac d{dt}\int_0^t f(s,t,a)ds=\int_0^t \frac {df(s,t,a)}{dt}+f(t,t,a)$$
which I tried to use to simplify your fractional derivative definition, but it created a division by $0$ error.
