A Functional Differential Equation: $f^\prime(x) =\frac{f(2x)}{2f(x)}$ I was having a play with some trig. identities and noticed the following:

$$\cos{x}=\frac{\sin{2x}}{2\sin{x}}.\tag{1}$$

Now, $\cos{x} = \frac{d}{dx}\sin{x}$ so I made the following analogous  differential equation:


  $$f^\prime(x) =\frac{f(2x)}{2f(x)} \tag{2}$$


I have not seen a differential equation which relates a function's derivative to a change in its argument, so I was wondering whether anyone knew what these were called?

Somewhat predictably, $f_1(x)=\sin{x}$ is not the only solution, I found that $f_2(x)=A\sin(\omega x)$ where $a\omega=1$ is also a solution. I then guessed another solution, $e^{\lambda x}$, and found that $f_3(x)=e^{\frac{1}{2}x}$ is also a solution.

My main questions are: 


  What are these type of equations called? and are there any other solutions to this one?


Thanks for reading.
CLAIM: The only solutions $f(x)$ to your FDE that are analytic at $x=0$ with $f(0)\ne 0$ are in the form $f(x)=ae^{x/2a}$ with $a\in\mathbb R$.


It is easy to demonstrate that functions of the described form satisfy your FDE, so all that remains is to establish that they are the only solutions.

Let's rewrite your functional equation as
$$2f(x)f'(x)=f(2x)$$
...ignoring the possible issues that this may cause when $f(x)=0$. Now let's assume that $f$ is analytic and that it has the Maclaurin series
$$f(x)=\sum_{n=0}^\infty \frac{x^n a_n}{n!}$$
for some coefficients $a_n$. Then this functional equation is the same as
$$2\bigg(\sum_{m=0}^\infty \frac{x^m a_m}{m!}\bigg)\bigg(\sum_{n=0}^\infty \frac{x^n a_{n+1}}{n!}\bigg)=\sum_{k=0}^\infty \frac{x^k 2^k a_{k}}{k!}$$
If we multiply together the series on the left, we get
$$\begin{align}
2\bigg(\sum_{m=0}^\infty \frac{x^m a_m}{m!}\bigg)\bigg(\sum_{n=0}^\infty \frac{x^n a_{n+1}}{n!}\bigg)
&amp;=\sum_{m,n=0}^\infty \frac{x^{m+n}\cdot 2 a_m a_{n+1}}{m!n!}\\
&amp;=\sum_{k=0}^\infty x^k\sum_{m+n=k}\frac{2a_m a_{n+1}}{m!n!}\\
&amp;=\sum_{k=0}^\infty x^k\sum_{m=0}^k \frac{2a_m a_{k-m+1}}{m!(k-m)!}\\
\end{align}$$
So our initial equality becomes
$$\sum_{k=0}^\infty x^k\sum_{m=0}^k \frac{2a_m a_{k-m+1}}{m!(k-m)!}=\sum_{k=0}^\infty x^k \frac{2^k a_{k}}{k!}$$
Now, by equating coefficients, we have that
$$\sum_{m=0}^k \frac{2a_m a_{k-m+1}}{m!(k-m)!}=\frac{2^k a_{k}}{k!}$$
...for all $k\in \mathbb Z$. This can be interpreted as a recurrence relation for $a_n$, and it demonstrates that given $a_0=f(0)$, all other coefficients, and thus the function $f$, are determined.

Thus, since for any $a_0$, there exists $a\in\mathbb R$ such that $f(x)=ae^{x/2a}$ satisfies $f(0)=a_0$, we have that the value of $f(0)$ determines the function $f(x)$. More specifically,
$$f(x)=f(0)e^{x/2f(0)}$$

NOTE. Because I began this strategy by multiplying both sides of your FDE by $f(x)$, this recurrence messes up if $f(0)=0$. Indeed, if this is the case, then the recurrence shows that $a_n=0$ for all $n$, and $f(x)=0$, even though you have found a function $f_2(x)=\sin(x\omega)/\omega$ that also satisfies $f_2(0)=0$ and is not constant. However, if $f(0)\ne 0$, the recurrence should work out just fine.
To use "Picard's method" might work. You have
$$
y(x)=y_0+\int_{x_0}^x \frac{y(2t)}{y(t)} \, dt
$$

Define 
$$
y_{k+1}=y_0+\int_{x_0}^x \frac{y_k(2t)}{y_k(t)}\,dt
$$
and calculate $\lim_{k\rightarrow \infty} y_k$.
I do not  know if the sequence converges.