"Differential equation" but we are solving for the order of integration I know and I have studied some differential equation and, normally, we want to solve for a function or a family of functions that do indeed solve the equation for example:
$$y'' - y' = y \sin(x)$$
But What if, instead of this, we have an equation and we want to solve for the order of the derivative? For example:
$$\frac{d^n}{dx^n} e^x = e^x$$
It is clear that in this particular case that the solution is the natural numbers, as any $n \in \Bbb N$ is the answer, but the thing is: this is just a guess. How do I know that this is the full answer. What I want to know is: What kind of mathematical tools and methods are there to solve this kinds of equations? We can generalize this even further and allow fractional derivatives and then get, for example:
$$\frac{d^r}{dx^r} e^x = e^x$$
Where $r$ is a real number. We can generalize this even further and get something like this:
$$\frac{d^r}{dx^r} y = y$$
Where in this case the solution will take form of $(r,y_r(x))$, because for each different value of $r \in \Bbb R$ there will be different functions $y_r$ that will satisfy this equation.
What kind of mathematical tools and methods are there to solve this kinds of equations?
 A: Since this is too long for a comment I'll make it an answer instead. Using the formula given in the comments and assuming $0<\alpha<1$, one can simplify it as
$$\mathrm{D}_x^\alpha f=\frac{1}{\Gamma(1-\alpha)}\int_0^x \frac{f'(u)}{(x-u)^\alpha}\mathrm{d}u$$
So,
$$\mathrm{D}_x^\alpha x=\frac{1}{\Gamma(1-\alpha)}\int_0^x \frac{1}{(x-u)^\alpha}\mathrm{d}u$$
Let $x-u=v\implies\mathrm{d}u=-\mathrm{d}v$. Then
$$\int_0^x\frac{1}{(x-u)^\alpha}\mathrm{d}u=\int_x^0v^{-\alpha}(-\mathrm{d}v)=\int_0^xv^{-\alpha}\mathrm{d}v=\frac{x^{1-\alpha}}{1-\alpha}$$
So
$$\mathrm{D}_x^\alpha x=\frac{x^{1-\alpha}}{(1-\alpha)\cdot\Gamma(1-\alpha)}$$
Using the recurrence formula for the Gamma function
$$\mathrm{D}_x^\alpha x=\frac{x^{1-\alpha}}{\Gamma(2-\alpha)}$$Therefore the equation $\mathrm{D}_x^\alpha x=\sqrt{x}$ has no solutions for $0<\alpha<1$ since
$$\mathrm{D}_x^{1/2}=\frac{\sqrt{x}}{\Gamma(3/2)}=2\sqrt{\frac{x}{\pi}}\neq\sqrt{x}.$$
EDIT: The above assumes what is known as the Caputo fractional derivative. If one instead uses the Riemann-Liouville fractional derivative,
$$\mathrm{D}_x^\nu f=\frac{1}{\Gamma(n-\nu)}\frac{\mathrm{d}^n}{\mathrm{d}x^n}\int_0^x (x-u)^{n-\nu-1}f(u)\mathrm{d}u$$
Then assuming $0<\alpha<1$,
$$\mathrm{D}_x^\alpha f=\frac{1}{\Gamma(1-\alpha)}\frac{\mathrm{d}}{\mathrm{d}x}\int_0^x (x-u)^{-\alpha}f(u)\mathrm{d}u$$
We can see that
$$\mathrm{D}_x^\alpha x=\frac{1}{\Gamma(1-\alpha)}\frac{\mathrm{d}}{\mathrm{d}x}\int_0^x\frac{u}{(x-u)^\alpha}\mathrm{d}u$$
Use $v=x-u\implies u=x-v \text{ and } \mathrm{d}u=-\mathrm{d}v$:
$$\int_0^x\frac{u}{(x-u)^\alpha}\mathrm{d}u=\int_{0}^{x}\frac{x-v}{v^\alpha}\mathrm{d}v=\left(x\frac{v^{1-\alpha}}{1-\alpha}-\frac{v^{2-\alpha}}{2-\alpha}\right)\bigg|_0^x=\frac{x^{2-\alpha}}{(1-\alpha)(2-\alpha)}$$
Therefore
$$\mathrm{D}_x^\alpha x=\frac{1}{\Gamma(1-\alpha)}\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^{2-\alpha}}{(1-\alpha)(2-\alpha)}\right)$$
$$=\frac{x^{1-\alpha}}{(1-\alpha)\Gamma(1-\alpha)}=\frac{x^{1-\alpha}}{\Gamma(2-\alpha)}$$
In this example the two methods return the same result, but this is not always the case.
