All the solutions of $f'(x)=f(x+\pi/2)$ Consider the following equation (with $f \in C^{\infty}(\mathbb{R})$): $$f'(x)=f(x+\pi/2)$$ 
This equation is satisfied by $f(x) = A\cos(x) +B\sin(x)$, for any $A,B \in \mathbb{R}$.
Question: What are all the (other) solutions of this equation (if any)?
 A: The "Ansatz" $f(x):=e^{\lambda x}$ leads to the equation
$$\lambda=e^{\lambda\pi/2}$$
with the obvious solutions $\lambda=\pm i$. But there are (probably an infinity)  other complex solutions, one of them being
$$a\pm ib:=1.0214 \pm 4.86821 \, i$$
(found numerically). The functions
$$f(x):=e^{ax}\bigl(A\cos(bx)+B\sin(bx)\bigr)$$
are then new solutions of your delay-differential equation.
A: A heuristic comment. The ansatz above works because the delay operator $f(\cdot) \mapsto f(\cdot+a)$ can be formally written as $e^{aD}$, where $D = \frac{d}{dx}$. This is simply the operator-side manifestation of the Taylor's theorem.
So your equation can be formally written as $(D - e^{\frac{\pi}{2}D}) f = 0$. Considering that the eigenfunctions of $D$ are exponential functions, this operator diagonalizes over the space of exponentials:
$$(\lambda - e^{\frac{\pi}{2}\lambda}) e^{\lambda x} = 0.$$
So it suffices to hunt for values $\lambda$ for which $\lambda - e^{\frac{\pi}{2}\lambda} = 0$ holds. Of course, all this discussion is not so rigorous and the possibility of other form of solutions is still open.
(Though I humbly suspect that the span of the solutions of the form above is in some sense 'dense in $C^{\infty}$' so that any solution is a possibly infinite linear combination of the solutions of the form above.)
