ODE problem with a single function but two arguments I have been trying to solve the following ODE with no success:
$$ \frac{df(x)}{dx} = -x f(x) + 4xf(2x)$$
I even tried using Maple but it seems to only accept ODE's that are function of the same argument (e.g., f(x) not f(2x)).
Any help would be appreciated!
 A: I'm afraid this is a type of so-called Functional Differential Equation, a notoriously difficult type of differential equations. The source of the difficulties is that only knowing an initial condition is not enough to `evolve' the function using the differential equations. To get a flavour of the difficulties, search for 'delay differential equations', which are actually easier than the one above. 
As an example, look at the very simple differential equation
\begin{equation}
 \frac{\text{d} f}{\text{d} x}(x) = f (2x).
\end{equation}
Suppose we can at least in a neighbourhood of $x=0$ expand a solution $f(x)$ as a power series in $x$, i.e. substitute $f(x) = \sum_{n=0}^\infty a_n x^n$. This gives the recurrence relation for $a_n$
\begin{equation}
 (n+1) a_{n+1} = 2^n a_n,
\end{equation}
which can be solved to obtain
\begin{equation}
 a_n = a_0 \frac{2^{\frac{1}{2} n(n-1)}}{\Gamma(1+n)}.
\end{equation}
However, the series
\begin{equation}
 \sum_{n=0}^\infty  a_0 \frac{2^{\frac{1}{2} n(n-1)}}{\Gamma(1+n)} x^n
\end{equation}
does not converge! So, it seems we cannot even write a solution $f(x)$ as a power series around $x=0$.
