Solve FDE: $f^{(1.5)}(x)=f(x)$

Question

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:

$e^x+e^{e^{(4/3)i\pi}x}$ ???!!

Answer 1

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)