No typo here.
I have been reading some articles on fractional order DE and most of them were engineering journals about numerical solutions (Please correct me if I am wrong).
$f'=f$ yields solutions in the form of $e^x$
$f''=f$ yields solutions in the form of $e^x+e^{-x}$
$f^{(1.5)}(x)=f(x)$ yields solutions in form of:
$e^x+e^{e^{(4/3)i\pi}x}$ ???!!
The most important issues for me are about e.g. the concept of the number of terms with arbitrarily constant.
Let $g(x)=f^{(0.5)}(x)$ ,
Then $g'''(x)-g''(x)=0$
$g(x)=C_1e^x+C_2x+C_3$ , which has well concept of $3$ terms with arbitrarily constant.
The issues of rational number order derivatives type fractional DEs VS irrational number order derivatives type fractional DEs (since the former can convert to ODE but the latter cannot)
That's a good question. This reminds me that in 11th grade I once gave a student presentation with a Powerpoint presentation (actually just about something very similar).
An Ordinary Differential Equation (ODE) containing differential operators of non-integer order is called a Fractional Differential Equation (FDE). (see ordinary-differential-equations and fractional-differential-equations)
Solving these FDEs is a lot more difficult than that of ordinary ODE, since they can have different solutions due to their fractional differential operators. This is a classic behavior in Fractional Calculus (see fractional-calculus) that there is no such thing as "the one Fractional Derivative". These Fractional Derivatives differ depending on the definition of the associated differential operator but also the derivation of the Fractional Derivatives. So what you mention is just one possible solution.
In the following I will use $\operatorname{D_{x}^{\alpha}}$ as a differential operator of the form $\operatorname{D_{x}^{\alpha}} = \frac{\operatorname{d}^{\alpha}}{\operatorname{d}x^{\alpha}}$ aka $f^{\left( \alpha \right)}\left( x \right) = \operatorname{D_{x}^{\alpha}}\left[ f\left( x \right) \right]$. This means that $\operatorname{D_{x}^{\alpha}}\left[ f\left( x \right) \right]$ is the $\alpha$th derivative with respect to $x$ of the function $f$ dependent on $x$.
Note: This implies $\int f\left( x \right)\, \operatorname{d}x = \operatorname{D_{x}^{-1}}\left[ f\left( x \right) \right]$ and $f\left( x \right) = \operatorname{D_{x}^{0}}\left[ f\left( x \right) \right]$.
If the differential equation is an FDE, it is homogeneous and linear, which allows us to reduce it to its characteristic equation with the substitution $f\left( x \right) = e^{\lambda \cdot x}$ and assuming that $\operatorname{D_{x}^{\alpha}}\left[ e^{\lambda \cdot x} \right] = \lambda^{\alpha} \cdot e^{\lambda \cdot x}$ still holds (not entirely true, but let's assume it anyway). The assumption allows us to solve it: $$ \begin{align*} \operatorname{D_{x}^{1.5}}\left[ f\left( x \right) \right] &= f\left( x \right)\\ \operatorname{D_{x}^{1.5}}\left[ e^{\lambda \cdot x} \right] &= e^{\lambda \cdot x}\\ \lambda^{1.5} \cdot e^{\lambda \cdot x} &= e^{\lambda \cdot x}\\ \lambda^{1.5} &= 1\\ \lambda &= \sqrt[1.5]{1}\\ \lambda &= 1^{\frac{1}{1.5}}\\ \lambda &= 1^{\frac{1}{\frac{3}{2}}}\\ \lambda &= 1^{\frac{2}{3}}\\ \lambda &= \left( 1 \right)^{\frac{2}{3}}\\ \lambda_{k} &= \left( e^{2 \cdot k \cdot \pi \cdot i} \right)^{\frac{2}{3}}\\ \lambda_{k} &= e^{\frac{2}{3} \cdot 2 \cdot k \cdot \pi \cdot i}\\ \lambda_{k} &= e^{\frac{4}{3} \cdot k \cdot \pi \cdot i}\\ \lambda_{k} &= \cos\left( \frac{4}{3} \cdot k \cdot \pi \right) + \sin\left( \frac{4}{3} \cdot k \cdot \pi \right) \cdot i\\ \end{align*} $$ Note: I used Euler's Formula for simplification and solving.
This gives us the roots of the characteristic equation ($\lambda_{3} = \underbrace{1}_{\Re\left( \lambda_{3} \right)} + \underbrace{0}_{\Im\left( \lambda_{3} \right)} \cdot i$ and $\lambda_{1,\, 2} = \underbrace{-0.5}_{\Re\left( \lambda_{1,\, 2} \right)} + \underbrace{\pm \frac{\sqrt{3}}{2}}_{\Im\left( \lambda_{1,\, 2} \right)} \cdot i$ aka the multiplicity of all $3$ roots $= 1$).
Via using $y\left( x \right) = Q_{0}\left( x \right) \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + P_{0}\left( x \right) \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right)$ aka $y\left( x \right) = Q_{0}\left( x \right) \cdot e^{\Re\left( \lambda_{k} \right) \cdot x} \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + P_{0}\left( x \right) \cdot e^{\Re\left( \lambda_{k} \right) \cdot x} \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right)$ we'll get: $$ \begin{align*} y\left( x \right) &= \sum\limits_{k = 1}^{3}\left[ c_{2 \cdot \left( k - 1 \right)} \cdot e^{\Re\left( \lambda_{k} \right) \cdot x} \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + c_{2 \cdot \left( k - 1 \right) + 1} \cdot e^{\Re\left( \lambda_{k} \right) \cdot x} \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right) \right]\\ y\left( x \right) &= c_{3} \cdot e^{x} + \sum\limits_{k = 1}^{2}\left[ c_{2 \cdot \left( k - 1 \right)} \cdot e^{\Re\left( \lambda_{k} \right) \cdot x} \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + c_{2 \cdot \left( k - 1 \right) + 1} \cdot e^{\Re\left( \lambda_{k} \right) \cdot x} \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right) \right]\\ y\left( x \right) &= c_{3} \cdot e^{x} + \sum\limits_{k = 1}^{2}\left[ c_{2 \cdot \left( k - 1 \right)} \cdot e^{-0.5 \cdot x} \cdot \cos\left( \pm\frac{\sqrt{3}}{2} \cdot x \right) + c_{2 \cdot \left( k - 1 \right) + 1} \cdot e^{-0.5 \cdot x} \cdot \sin\left( \pm\frac{\sqrt{3}}{2} \cdot x \right) \right]\\ y\left( x \right) &= c_{3} \cdot e^{x} + \sum\limits_{k = 1}^{2}\left[ \left( c_{2 \cdot \left( k - 1 \right)} \cdot \cos\left( \frac{\sqrt{3}}{2} \cdot x \right) \pm c_{2 \cdot \left( k - 1 \right) + 1} \cdot \sin\left( \frac{\sqrt{3}}{2} \cdot x \right) \right) \cdot e^{-0.5 \cdot x} \right]\\ y\left( x \right) &= c_{3} \cdot e^{x} + e^{-0.5 \cdot x} \cdot \sum\limits_{k = 1}^{2}\left[ c_{2 \cdot \left( k - 1 \right)} \cdot \cos\left( \frac{\sqrt{3}}{2} \cdot x \right) \pm c_{2 \cdot \left( k - 1 \right) + 1} \cdot \sin\left( \frac{\sqrt{3}}{2} \cdot x \right) \right]\\ \end{align*} $$ where all $c$s are arbitrarily constants.
You can combine all of this into your solution by saying that the arbitrarily constant factors of $\sin\left( \cdot \right)$ are $= -i$ and all other arbitrarily constant factors are $= 1$: $$ \begin{align*} y\left( x \right) &= c_{3} \cdot e^{x} + e^{-0.5 \cdot x} \cdot \sum\limits_{k = 1}^{2}\left[ c_{2 \cdot \left( k - 1 \right)} \cdot \cos\left( \frac{\sqrt{3}}{2} \cdot x \right) \pm c_{2 \cdot \left( k - 1 \right) + 1} \cdot \sin\left( \frac{\sqrt{3}}{2} \cdot x \right) \right]\\ y\left( x \right) &= 1 \cdot e^{x} + e^{-0.5 \cdot x} \cdot \sum\limits_{k = 1}^{2}\left[ 1 \cdot \cos\left( \frac{\sqrt{3}}{2} \cdot x \right) - i \cdot \sin\left( \frac{\sqrt{3}}{2} \cdot x \right) \right]\\ y\left( x \right) &= e^{x} + e^{-0.5 \cdot x} \cdot \sum\limits_{k = 1}^{2}\left[ \cos\left( \frac{\sqrt{3}}{2} \cdot x \right) - i \cdot \sin\left( \frac{\sqrt{3}}{2} \cdot x \right) \right]\\ y\left( x \right) &= e^{x} + e^{-0.5 \cdot x} \cdot \sum\limits_{k = 1}^{2}\left[ e^{-\frac{\sqrt{3}}{2} \cdot x \cdot i} \right]\\ y\left( x \right) &= e^{x} + e^{-0.5 \cdot x} \cdot e^{-\frac{\sqrt{3}}{2} \cdot x \cdot i}\\ y\left( x \right) &= e^{x} + e^{-0.5 \cdot x} \cdot e^{-\frac{\sqrt{3}}{2} \cdot x \cdot i}\\ y\left( x \right) &= e^{x} + e^{-0.5 \cdot x - \frac{\sqrt{3}}{2} \cdot x \cdot i}\\ y\left( x \right) &= e^{x} + e^{\left( -0.5 - \frac{\sqrt{3}}{2} \cdot i \right) \cdot x}\\ y\left( x \right) &= e^{x} + e^{e^{\frac{4}{3} \cdot \pi \cdot i} \cdot x}\\ \end{align*} $$ Note: I used the simplification that we know as a root of the characteristic equation ($-0.5 - \frac{\sqrt{3}}{2} \cdot i = \lambda_{1} = e^{\frac{4}{3} \cdot \pi \cdot i}$) for solving.
So the general solution is: $$\fbox{$ \begin{align*} y\left( x \right) &= c_{3} \cdot e^{x} + e^{-0.5 \cdot x} \cdot \sum\limits_{k = 1}^{2}\left[ c_{2 \cdot \left( k - 1 \right)} \cdot \cos\left( \frac{\sqrt{3}}{2} \cdot x \right) + c_{2 \cdot \left( k - 1 \right) + 1} \cdot \sin\left( \frac{\sqrt{3}}{2} \cdot x \right) \right]\\ \end{align*} $} \tag{1}$$
Sometimes differential operators provide useful definitions and formulas that can be used to convert the FDE into an ODE / integral equation, which is sometimes easier to solve.
The Riemann–Liouville Fractional Operator, on the other hand, uses the formulas for simplifying mutible integrals into single integral:: $$ \begin{align*} \operatorname{_{a}D^{-v}_{x}}\left[ f\left( x \right) \right] \equiv \frac{1}{\Gamma\left( v \right)} \cdot \int\limits_{a}^{x} f\left( u \right) \cdot \left( x - u \right)^{v - 1}\, \operatorname{d}u \tag{*}\\ \end{align*} $$ where $\Gamma\left( \cdot \right)$ ist the Complte Gamma Function.
Using this definition, we can rewrite the ODE as follows: $$ \begin{align*} \operatorname{_{a}D_{x}^{1.5}}\left[ f\left( x \right) \right] &= f\left( x \right)\\ \frac{1}{\Gamma\left( -1.5 \right)} \cdot \int\limits_{a}^{x} f\left( u \right) \cdot \left( x - u \right)^{-1.5 - 1}\, \operatorname{d}u &= f\left( x \right)\\ \frac{1}{\Gamma\left( -1.5 \right)} \cdot \int\limits_{a}^{x} f\left( u \right) \cdot \left( x - u \right)^{-2.5}\, \operatorname{d}u &= f\left( x \right)\\ \frac{1}{\Gamma\left( -1.5 \right)} \cdot \int\limits_{a}^{x} \frac{f\left( u \right)}{\left( x - u \right)^{2.5}}\, \operatorname{d}u &= f\left( x \right)\\ \end{align*} $$
So you would get: $$ \begin{align*} \frac{1}{\Gamma\left( -1.5 \right)} \cdot \int\limits_{a}^{x} \frac{f\left( u \right)}{\left( x - u \right)^{2.5}}\, \operatorname{d}u &= f\left( x \right) \tag{**}\\ \end{align*} $$
But I will leave the solving of this integral equation / the ODE to you as a small exercise.
If you get stuck here, you could also try the whole thing with the convolution formula.
Another popular method for solving FDEs is manipulating the order of the FDE: We can try to substitute the function in such a way that we get a new intiger-order differential equation (an ODE) and solve it with the known methods or the order of the differential operators themselves substitute that a new FDE comes out, which is easier to solve or easier to simplify.
We know that in calculus $\operatorname{D_{x}^{m}}\left[ \operatorname{D_{x}^{n}}\left[ f\left( x \right) \right] \right] = \operatorname{D_{x}^{m + n}}\left[ f\left( x \right) \right]$ holds for all $m \in \mathbb{Z} \ni n$. If we claim that this also applies to $m \in \mathbb{Q} \ni n$ we can rewrite the whole equation.
I won't do this here since @doraemonpaul allready tryied that.
At first rewriting: $$ \begin{align*} \operatorname{D_{x}^{1.5}}\left[ f\left( x \right) \right] &= f\left( x \right)\\ \operatorname{D_{x}^{2 \cdot \underbrace{0.75}_{v}}}\left[ f\left( x \right) \right] &= f\left( x \right)\\ \operatorname{D_{x}^{2 \cdot v}}\left[ f\left( x \right) \right] &= f\left( x \right)\\ \operatorname{D_{x}^{2 \cdot v}}\left[ f\left( x \right) \right] - f\left( x \right) &= 0\\ \operatorname{D_{x}^{2 \cdot v}}\left[ f\left( x \right) \right] + \underbrace{0}_{a} \cdot \operatorname{D_{x}^{v}}\left[ f\left( x \right) \right] + \underbrace{- 1}_{b} \cdot f\left( x \right) &= 0\\ \operatorname{D_{x}^{2 \cdot v}}\left[ f\left( x \right) \right] + a \cdot \operatorname{D_{x}^{v}}\left[ f\left( x \right) \right] + b \cdot f\left( x \right) &= 0\\ \end{align*} $$
This is an well studied DE and has a solution given by: $$\fbox{$ \begin{align*} y\left( x \right) &= \begin{cases} \sum\limits_{k = 0}^{\frac{1}{v} - 1}\left( a^{\frac{1}{v} - k - 1} \cdot \operatorname{E}_{x}\left( -k \cdot v;\, a^{\frac{1}{v}} \right) - b^{\frac{1}{v} - k - 1} \cdot \operatorname{E}_{x}\left( -k \cdot v;\, b^{\frac{1}{v}} \right) \right), &\text{if}\, a \ne b\\ x \cdot \exp\left( a \cdot x \right) \cdot \sum\limits_{k = -\frac{1}{v} + 1}^{\frac{1}{v} - 1}\left( a^{k} \cdot \left( \frac{1}{v} - \left| k \right| \right) \cdot \exp\left( a^{\frac{1}{v}} \cdot x \right) \right), &\text{if}\, a = b \ne 0\\ \frac{1}{\Gamma\left( 2 \cdot v \right)} \cdot x^{2 \cdot v - 1 }, &\text{if}\, a = b = 0\\ \end{cases}\\ \end{align*} $} \tag{3}$$ where $\Gamma\left( \cdot \right)$ is the Complete Gamma Function and $\operatorname{E}_{t}\left( \cdot;\, \cdot \right)$ is the $\operatorname{E}_{t}$-Function.
For reference see: Weisstein, Eric W. "Fractional Differential Equation." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/FractionalDifferentialEquation.html