Finding non-zero function such that $f(2x)=f'(x)\cdot f''(x)$ I am thankful if someone can help me or show me the clue. As honestly as possible, I got stuck on this problem.
I need some help in finding a $f(x)\neq 0$ such that $$f(2x)=f'(x) \cdot f''(x),$$ where $f',f''$ are the first and second derivatives, respectively. It is not an ordinary differential equation.
My last try was to put $f(x)=ke^{ax}$, and then
$$ke^{2ax}=kae^{ax}\cdot ka^2e^{ax}.$$ Now by canceling $e^{ax}$, we have $$k=k^2a^3.$$ This shows $k=1,a=1$ and finally $f(x)=e^{x}$. But I want to solve the problem analytically.
 A: Too long for a comment.
Assuming $f(x)$ is polynomial of degree $n$, we have $n = (n-1)+(n-2)$ or $n = 3$. Therefore we can look for $f$ in the following form:
$$\begin{align}
f(x) &= a{x^3} + b{x^2} + cx + d \\ 
f'(x) &= 3a{x^2} + 2bx + c \\ 
f''(x) &= 6ax + 2b \end{align}$$
Then,
$$f(2x) = f'(x) \cdot f''(x) \implies \begin{cases}8a = 18a^2 \\ 4b = 18ab \\ 2c = 4b^2 + 6ac \\
d = 2bc\end{cases}$$
which gives the solutions $(0,0,0,0)$ or $(4/9,0,0,0)$. So the only polynomial solution is $$f(x) = \frac{4}{9}x^3$$
A: As was observed in answers and comments, the only polynomial solution is $f(x)=\frac{4}{9}x^3$.
Your own work leads to the solutions $f(x) = \frac{1}{a^3}e^{a x}$.
Adding the initial conditions $f(0)=0$, $f'(0)\ne 0$ and assuming that $f$ is analytic, looking at the coefficients of the Taylor series we get $f(x) = \frac{2}{a^3} \sinh(a x)$
Without the initial conditions $f(0)=0$ and using the Taylor series it's clear that there's more solutions, but probably there's no simple formula for the coefficients of the series.
Without the assumption of the function been analytic, I have no idea how the problem could be attacked.
A: If $f$ is a solution and $a$ a positive number then $a^{-3}f(ax)$ is also a solution, as the examples so far illustrate.
A: Notice that $f'(x)\cdot{f''(x)}=\frac12(f'\cdot{f'})'(x)=f(2x),$ so that $(f'\cdot{f'})'(x)=2f(2x).$ Let $F$ be the antiderivative of $f$ with $F(0)=0.$ As such, let $F_2(x)=F(2x),$ so $F_2'(x)=2F'(2x)=2f(2x).$ Therefore, $f'(x)^2-f'(0)^2=F(2x).$ Perhaps the latter is more easily solvable for $f$ in terms of $F,$ and the equation can then latter be solved for $F$ when reformulated in terms of $F.$
