Solving $f'(t)=f(t+\pi/2)$ As the title indicates, I want to solve the first order ODE $f'(t)=f(t+\pi/2)$. Here, $f$ is a real function with real variable $t$, and the equation is true for all real $t$.
Immediately, I can tell that $f(t)=\sin t$ or $\cos t$ are solutions, since $f'(t)=\frac{d}{dt}\sin t=\cos t = \sin(t+\pi/2)=f(t+\pi/2)$ and similarly for the cosine. After some pondering, I realised that $f(t)=C_1\cos t + C_2\sin t$ is a solution as well for any constants $C_1$ and $C_2$. I'm tempted to think that these are all the solutions, but I cannot say so for sure; I know that these are all the solutions if the principle of superposition applies, but the way I learnt it, the principle only applies to differential equations of the form $p(D)f(t)=g(t)$ where $p$ is a polynomial.
Thus, I have two questions. Firstly, is what I have given indeed all of the solutions, and how do you prove that? (Perhaps equivalently, how do you prove or disprove that the principle of superposition holds for any imaginable differential equation, no matter what form it is in?) Secondly, and more importantly, how do you properly solve this question? I got my solutions based on pure guessing and not proper reasoning, but there has to be a method to solve this step-by-step without such a huge leap of faith! I have googled for this exact question, but as most resources out there are dedicated to solving equations where the argument of the functions do not change (i.e. the arguments are always $t$ and not $t+\pi/2$), I couldn't find anything useful. 
 A: There are other solutions, but I do not know whether I have obtained all possible solutions (most likely not).  For $k\in\mathbb{Z}$, let $W_k$ denote the $k$-th branch of the Lambert $W$-function (noting that $W_0$ is the principal branch).  We then set 
$$\alpha_k:=-\frac{2}{\pi}\,W_k\left(-\frac{\pi}{2}\right)\,.$$
In fact, we have
$$\alpha_{-1-k}=\bar{\alpha}_k\text{ for every }k\in\mathbb{Z}\,.$$
For example,
$$\alpha_{-1}=\text{i}\text{ and }\alpha_0=-\text{i}\,.$$
Now, define
$$\begin{align}f_k(x)&:=\exp\left(\alpha_k\,x\right)\\&=\exp\big(\text{Re}(\alpha_k)\,x\big)\,\Big(\cos\big(\text{Im}(\alpha_k)\,x\big)+\text{i}\,\sin\big(\text{Im}(\alpha_k)\,x\big)\Big)\end{align}$$
for each $x\in\mathbb{R}$.
Then, for any $k\in\mathbb{Z}$, the function $f_k$ satisfies
$$f'_k(x)=f_k\left(x+\frac{\pi}{2}\right)\text{ for every }x\in\mathbb{R}\,,$$
because
$$\exp\left(\frac{\pi}{2}\,\alpha_k\right)=\alpha_k\,.$$
Furthermore, any $\mathbb{C}$-linear combination $f$ of the functions $f_k$, where $k\in\mathbb{Z}$, obeys the differential equation
$$f'(x)=f\left(x+\frac{\pi}{2}\right)\text{ for all }x\in\mathbb{R}\,.$$
In particular,
$$\cos=\frac{1}{2}\,\alpha_{-1}+\frac{1}{2}\,\alpha_0$$
and
$$\sin=\frac{1}{2\text{i}}\,\alpha_{-1}-\frac{1}{2\text{i}}\,\alpha_0\,.$$
A: *

*Take any infinitely differentiable function $g(x)$ on $(0,\pi)$.

*Grab any infinitely smooth step function $s(x)$ that goes from 0 to 1. This will do: Example of a smooth 'step'-function that is constant below 0 and constant above 1

*Define $f(x)=g'(x)s(x)+g(x+{\pi\over2})(1-s(x))$ for $x\in(0,{\pi\over2})$. This way you'll make your derivative on the left end match your function on the right end.

*Define $f(x)$ for $x\in({\pi\over2},\pi)$ via your equation applied in reverse (that is, using $f'(x)$ to find $f(x+{\pi\over2})$).

*Continue to $(\pi,{3\pi\over2})$ and so on.


You may also continue the other way. That's much easier, since we don't have to bother about smoothness; integration will produce it for us.
Now it seems general enough.
