TITLE: Solution to the differential equation $G'(x) = G(\pi x)$
QUESTION [7 upvotes]: In this answer, I came across the differential equation
$$
G'(x) = G(\pi x)
$$
Despite its apparent simplicity, I couldn't think of an elementary solution to it. Does this have a known solution?

REPLY [3 votes]: Let me try a power series
$$
G(x) =\sum_{n=0}^{\infty}a_n x^n
$$
$$
G(\pi x)=\sum_{n=0}^{\infty}a_n\pi^n x^n
$$
$$
G'(x)=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n
$$
Setting $G'(x) = G(\pi x)$ we have the following recursion relation among the coefficients of the power series $G(x)$.
$$
a_{n+1}=\frac{\pi^n}{n+1}a_{n},\;\;\;n=0,1,\cdots
$$
$$
a_1=a_0
$$
$$
a_2=\frac{\pi}{2}a_1=\frac{\pi}{2!}a_0
$$
$$
a_3=\frac{\pi^2}{3}a_2=\frac{\pi^3}{3!}a_0
$$
$$
a_4=\frac{\pi^3}{4}a_3=\frac{\pi^6}{4!}a_0
$$
$$
a_n=\frac{\pi^{\sum_{j=1}^{n-1}j}}{n!}a_0
$$
$$
\sum_{j=1}^{n-1}j=\frac{n(n-1)}{2}
$$
$$
a_n=\frac{\pi^{n(n-1)/2}}{n!}
$$
$$
G(x)=a_0\sum_{n=0}^{\infty}\frac{\pi^{n(n-1)/2}}{n!}x^n
$$
$$
G(x)=a_0 G_0(x)
$$
where $G_0(x)$ is the solution of the above equation for which $G_0(0)=1$.
$$
G_0(x)=\sum_{n=0}^{\infty}\frac{\pi^{n(n-1)/2}}{n!}x^n
$$
I'm concerned about the convergence properties of the power series but this is my best attempt at the problem.