Solving Differential Functional Equation $f(2x)=2f(x)f'(x)$ Find all functions satisfying $f(2x)=2f'(x)f(x)$
Under given condition, can't we find explicit solutions?
 A: Assume that a solution $f(x)$ is analytic in some neighborhood of $x=0$ so that it can be represented as a converging power series
$$
f(x)=a_0+a_1x+a_2x^2+\cdots.
$$
If we assume that $a_0=f(0)\neq0$, then, by equating the power series representing the two sides of the given equation, we see that the remaining coefficients are uniquely determined. Comparing the coefficient of $x^\ell$ on both sides gives a linear equation for $a_{\ell+1}$ with a non-zero coefficient. 
So if an analytic solution with a fixed value $f(0)=k\neq0$ exists, it is unique. An ansatz of the form
$$f(x)=ke^{\lambda x}$$
as in rajb245's answer works, iff $\lambda=1/2k$. Therefore we have a one-parameter family of solutions
$$
f(x)=f_k(x)=ke^{x/2k}.
$$
If $a_0=0$ things become messy, as the proffered solutions $f(x)=x$ and $f(x)=\sin x$ prove. Unless I made a mistake, we get that then $a_1=0$ or $a_1=1$. In the latter case I got that we necessarily have $a_2=0$, but $a_3$ can be arbitrary, which goes together with the known solutions. Probably this approach can be taken further (most likely by somebody else)...

Again assuming analyticity, and taking advantage of mezhang's observation that the r.h.s. is the derivative of $f(x)^2$ we get by differentiating the given equation $k$ times we get the equations
$$
2^kf^{(k)}(2x)=D^{k+1}\left[f(x)^2\right]=\sum_{i=0}^{k+1}{k+1\choose i} f^{(i)}(x)f^{(k+1-i)}(x).
$$
Plugging in $x=0$ gives then the system of equations
$$
2^kf^{(k)}(0)=\sum_{i=0}^{k+1}{k+1\choose i} f^{(i)}(0)f^{(k+1-i)}(0)
$$
that holds for all $k$. As $f^{(k)}(0)=k!a_k$ we get
$$
2^k k! a_k=(k+1)!\sum_{i=0}^{k+1} a_ia_{k+1-i}.
$$
Assuming (see above) $a_0=0,a_1=1$ this gives us
$$
2^kk!a_k=(k+1)!\left(2a_k+\sum_{i=2}^{k-1}a_ia_{k+1-i}\right).
$$
With $k=2$ we get $a_2=0$ (confirming my hand calculations). With $k=3$ we get a tautology $48a_3=48a_3$ that does, indeed, leave the value of $a_3$ undetermined.
For $k>3$ we have $2^kk!>2(k+1)!$, so we get an equation recursively determining
$a_k$. Observe that a straightforward induction proves (after this start) that $a_k=0$ whenever $k$ is even. Thus the resulting function $f(x)$ is odd, and uniquely determined by the value of $a_3$. I don't know, if the solution is always an elementary function.
In the case $a_0=a_1=0$ the term $(k+1)!(a_1a_k+a_ka_1)=2(k+1)!a_ka_1$ is not there on the r.h.s., so we get the equations
$$
2^kk!a_k=(k+1)!\left(\sum_{i=2}^{k-1}a_ia_{k+1-i}\right)
$$
for all $k$. Again we recursively see that this system has only the trivial solution $a_k=0$ for all $k$.
A: Suppose the ansatz $f(x)=e^{kx}$, then $f(2x)=e^{2kx}=f(x)^2$, $f'(x)=ke^{kx}=kf(x)$. The equation is
$$
f^2(x)=2 (k f(x))f(x)
$$
$$
f^2(x)=2k f^2(x)\Rightarrow k=1/2
$$
Substituting that $k$ back in gives $f(x)=e^{x/2}$
Of course this doesn't seem to get at all the solutions...it seems $f(x)=x$ is another solution, but this is a start on the exponential solutions. You can look for power solutions using a similar method...
Edit
Users have mentioned the following are all solutions
\begin{align}
f(x)&=0 \\
f(x) &= x\\
f(x) &= \sin(x)\\
f(x) &= e^{x/2}
\end{align}
I don't know the technique to get the general solution to these types of equations.  I'm not even sure what to google.  "Functional differential equation" is moderately helpful...
