Solving a differential equation with a factor $2$ in the argument of the unknown function. The well-known trig identity 
$$2\sin(x)\cos(x) = \sin(2x)$$ 
gives use the differential equation (which I'm not even sure qualifies as an ODE, given the factor in the argument?)
$$2f(x)f'(x) = f(2x)$$
Assuming one had no idea that $\sin(x)$ is a solution to this equation, I was wondering if there was a way to derive the solution anyway.
My attempt
Assume that $f(x)$ has a power series expansion valid for all values of $x$ that we are interested in, that is
$$f(x) = \sum_{n=0}^{\infty}a_nx^n$$
for some coefficients $a_n$.
Plugging this in gives us
$$\sum_{n=0}^\infty 2^na_nx^n=2\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty (n+1)a_{n+1}x^n\right) \tag1$$
Assuming absolute convergence, we can rewrite the RHS using the Cauchy product formula:
\begin{align}
\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty (n+1)a_{n+1}x^n\right)&=\sum_{n=0}^\infty\sum_{k=0}^n(k+1)a_{k+1}x^ka_{n-k}x^{n-k}\\
&=\sum_{n=0}^\infty x^n\sum_{k=0}^n(k+1)a_{k+1}a_{n-k}
\end{align}
Plugging this into $(1)$ and comparing coefficients of powers of $x$, we get:
$$a_n=2^{1-n}\sum_{k=0}^n(k+1)a_{k+1}a_{n-k}$$
However, I have no idea how to proceed with this recurrence relation for the coefficients of the power series that I am searching for.

Therefore, my two questions are:
$1.$ Is my approach correct and is there a way to solve this recurrence relation for the known coefficients of the power series expansion of $\sin(x)?$
$2.$ Is there a general approach that is maybe less cumbersome to this kind of problem?

 A: One could imagine the solution to be an exponential of the following form:
$$f(x)=ae^{bx}$$
Substituting values in, we have
$$2a^2be^{2bx}=ae^{2bx}$$
$$\implies2ab=1$$
$$\implies a=1/2b$$
$$\implies f(x)=\frac1{2b}e^{bx}$$
which is another more general solution to your differential equation.  If $b$ is complex, we might end up with $f(x)=\sin(x)$ due to Euler's formula.
A: Entering the "Ansatz"
$$f(x):=\sum_{k=0}^n a_k x^k$$
with, e.g., $n=8$ into Mathematica and putting the  coefficients of
$$\psi:=2f(x)f'(x)-f(2x)+O[x]^{n+1}$$
to $0$ produces successive numerical values $a_k$ $(0\leq k\leq n)$ that can be attributed to the following solution functions:
$$f(x)=\qquad x\>,\qquad{1\over2\lambda}e^{\lambda x}\>,\qquad {1\over\lambda}\sin(\lambda x)$$
with $\lambda\ne0$ an arbitrary complex constant.
Note that a solution which is $\ne0$ in a punctured neighborhood $\dot U$ of $0$ is automatically $C^\infty$ in $\dot U$. But there might exist solutions that have an essential singularity at $0$.
