TITLE: Solving Differential Functional Equation $f(2x)=2f(x)f'(x)$
QUESTION [16 upvotes]: Find all functions satisfying $f(2x)=2f'(x)f(x)$
Under given condition, can't we find explicit solutions?

REPLY [12 votes]: If one restrict oneself to functions which admit a power series expansion at $x = 0$,
following is the list of solutions I found:
$$f(x) = \begin{cases}
0,             & f(0) = 0, f'(0) = 0,\\
k \sin(x/k),   & f(0) = 0, f'(0) = 1, f'''(0) < 0\\
x,             & f(0) = 0, f'(0) = 1, f'''(0) = 0\\
k \sinh(x/k),  & f(0) = 0, f'(0) = 1, f'''(0) > 0\\
k \exp(x/2k),  & f(0) \ne 0, f'(0) = \frac12\\
\end{cases}$$
where $k$ is a non-zero constant.
Let $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$. In terms of $a_n$, the function equation becomes:
$$\begin{align}
a_0     &= 2 ( a_1 a_0 )\tag{*0}\\
2\,a_1  &= 2 ( 2 a_2 a_0 + a_1 a_1 )\tag{*1}\\
2^2 a_2 &= 2 ( 3 a_3 a_0 + 2 a_2 a_1 + a_1 a_2 )\\
2^3 a_3 &= 2 ( 4 a_4 a_0 + 3 a_3 a_1 + 2 a_2 a_2 + a_1 a_3 )\\
2^4 a_4 &= 2 ( 5 a_5 a_0 + 4 a_4 a_1 + 3 a_3 a_2 + 2 a_2 a_3 + a_1 a_4)\\
2^5 a_5 &= 2 ( 6 a_6 a_0 + 5 a_5 a_1 + 4 a_4 a_2 + 3 a_3 a_3 + 2 a_2 a_4 + a_1 a_5 )\\
&\;\vdots
\end{align}$$
The first thing we notice is $(*0)$ implies:
$$a_0 (1 - 2a_1) = 0 \implies a_0 = 0 \quad\text{ or }\quad a_1 = \frac12$$
If $a_0 = 0$, then $(*1)$ implies $a_1 = 0$ or $1$. It is easy to see $a_1 = 0$ will lead us to $f(x) \equiv 0$. So let us assume $a_1 = 1$. The set of equations become:
$$\begin{align}
( 2^1 - 3 ) a_2 &= 0\\
( 2^2 - 4 ) a_3 &= 2 a_2 a_2 \iff 0 = 0 \text{ ( i.e. } a_3\;\text{ is arbitrary )}\\
( 2^3 - 5 ) a_4 &= 3 a_3 a_2 + 2 a_2 a_3\\
( 2^4 - 6 ) a_5 &= 4 a_4 a_2 + 3 a_3 a_3 + 2 a_2 a_4\\
&\;\vdots
\end{align}$$
It is then easy to see:

$a_n = 0$ for all even $n$.
$a_3$ is arbitrary.
once $a_3$ is fixed, $a_n$ is recursively fixed for all odd $n > 3$.

Depends on the sign of $a_3$, it leads us to solutions of the form $k \sin(x/k), x $ or k sinh(x/k)$.
If $a_0 \ne 0$, then $a_1 = \frac12$ and the set of equations becomes:
$$\begin{align}
2 a_2 a_0 &= 2^0\,a_1   - ( a_1 a_1 )\tag{*1}\\
3 a_3 a_0 &= 2^1 a_2  - ( 2 a_2 a_1 + a_1 a_2 )\\
4 a_4 a_0 &= 2^2 a_3 - ( 3 a_3 a_1 + 2 a_2 a_2 + a_1 a_3 )\\
5 a_5 a_0 &= 2^3 a_4  - ( 4 a_4 a_1 + 3 a_3 a_2 + 2 a_2 a_3 + a_1 a_4)\\
&\;\vdots
\end{align}$$
It is easy to see these will recursively fix the values of $a_n$ for all $n > 1$. This lead us to the $k \exp(x/2k)$ solution.