Power series solution to a differential equation If $f(\rho)$ satisfies $\frac{df}{d\rho}=\frac{f(2\rho)}{2f(\rho)}$, I am trying to derive the form of $f(\rho)$ by using a power series expansion $f(\rho)=\sum a_n \rho^n$ and show that $f(\rho)$ can be $\rho$, $R\sin(\rho/R)$ or $R\sinh(\rho/R)$. I am getting stuck.
What should be further steps?
Thanks for the help in advance.
 A: This is an interesting problem, albeit not an ODE in the usual sense. I don't think there is a general theory for equations of your kind. It turns out that in your example there are various families of elementary solutions (you found some yourself); but this seems a happy coincidence.
The solutions $t\mapsto f(t)$ which are analytic in a neighborhood of $t=0$ can be found using a power series "Ansatz" $$f(t)=\sum_{k=0}^\infty a_kt^k\tag{1}$$ and carefully analyzing the consequences of arbitrary choices made during the first steps. Plugging this "Ansatz" into the equation
$$2f(t)f'(t)=f(2t)\tag{2}$$ we obtain
$$2\sum_{j\geq0, \>k\geq0}(j+1)a_{j+1}a_k t^{j+k}-\sum_{r\geq0}2^r a_r t^r\equiv0\ .$$
Collecting terms belonging to $t^r$ leads to the equations
$$2\sum_{j+k=r}(j+1)a_{j+1}a_k-2^r a_r=0\qquad(r\geq0)\ .$$
The first two equations are
$$
a_0(2a_1-1)=0\>,\qquad
2a_0a_2=a_1(1-a_1)\ .$$
The first equation enforces (i) $a_0=0$ or (ii) $a_1={1\over2}$. 
In case (i) the second equation enforces (i.i) $a_1=0$ or (i.ii) $a_1=1$. In case (i.i) it will later turn out that all $a_k=0$. The resulting $f$ solves $(2)$, but not the original equation in the question. In case (i.ii) it will later turn out that all $a_k$ with even $k$ are $=0$, and that $a_3:=c$ is arbitrary. The remaining $a_k$ with odd $k$ are then determined.
In case (ii) we may choose $a_0:=c$ arbitarily, and it will then turn out that all $a_k$ are uniquely determined.
It comes as a surprise that in all cases we can recognize the found coefficients as coefficients of well known functions, and it is then easy to verify that these functions indeed solve the  equation $(2)$. In this way one obtains the solutions $$f_0(t)=t, \quad f_1(t):=\sin t, \quad f_2(t)=\sinh t,\quad f_3(t)=e^{t/2}\ ,$$ whereby in each case the parameter $c$ has been suppressed here: It follows from the special form of the given functional equation that when ${\cal G}(f)$ is the graph of a solution $f$ then stretching this graph from $(0,0)$ by a factor $R>0$ results in the graph of a solution as well. 
But it is still possible that someone will come up with other solutions which are $C^\infty$ at $t=0$, but not analytic there.
