I would like to find the solution of the imaginary order differential equation $y^{(2i)}+y^{(i)}+y=0$
I started with the Fourier transform differintegral as it seemed more suitable than the Riemann-Liouville fractional derivative, as I saw on this post: (Imaginary-Order Derivative): $$f^{(s)}(x)=\dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{+\infty} e^{- i \omega x}(-i \omega)^s \displaystyle\int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega.$$ Then I applied it to the equation, including $y(x)=y^{(0)}(x)$: $$f^{(s)}(x)=\dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \displaystyle\int_{-\infty}^{\infty}e^{i\omega(t-x)}y(t)(-i\omega)^{s}dtd\omega,$$ and got to $$\dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \displaystyle\int_{-\infty}^{\infty}e^{i\omega(t-x)}y(t)(-i\omega)^{2i}dtd\omega + \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \displaystyle\int_{-\infty}^{\infty}e^{i\omega(t-x)}y(t)(-i\omega)^{i}dtd\omega +\dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \displaystyle\int_{-\infty}^{\infty}e^{i\omega(t-x)}y(t)(-i\omega)^{0}dtd\omega$$
$$\dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \displaystyle\int_{-\infty}^{\infty}e^{i\omega(t-x)}y(t)((-i\omega)^{2i}+(-i\omega)^{i}+(-i\omega)^{0})dtd\omega=0.$$
taking the derivative of both sides doesn't help to get any solutions besides $y=0$.
Any help would be appreciated.