What type of differential equation is $f'(x) = f(x/2)$ and how do you solve it? I have the following different equation
$$f'(x) = f(x/2)$$
with $f(0)=10$.
What type of DE is this, and how would you solve it?
It seems $f(x)$ is likely to be some relative of $e^x$, since $f'(x) = f(x)$, which is close, but I don't even know what that type of DE is called with that $f(x/2)$ feature, so I'm not having any luck searching for a tutorial. The best candidate I've found was a "delay differential equation", but that seems more suited to $f(x-3)$ than $f(x/2)$.
 A: Notice
$${f'(x)=f\left(\frac{x}{2}\right)\Leftrightarrow f''(x) = \frac{1}{2}f'\left(\frac{x}{2}\right), f'''(x)=\frac{1}{2^2}f''\left(\frac{x}{2}\right)}$$
Plugging in some numbers we get ${f(0) = 10, f'(0) = 10, f''(0) = \frac{10}{2}, f'''(0) = \frac{1}{2^2}\left(\frac{10}{2}\right)=\frac{10}{2^3},f''''(0) = \frac{10}{2^6}}$... In general we see
$${f^{(n)}(0)=\frac{10}{2^{\left(\frac{n(n-1)}{2}\right)}}}$$
And now we can create a power series solution using a taylor series centered around $0$:
$${f(x)=10 + \sum_{n\geq 1}\frac{10x^n}{2^{\left(\frac{n(n-1)}{2}\right)}n!}}$$
A: If you let:
$x=2^t$
then remember we have:
$dx=\ln(2)2^tdt$ and our original equation is:
$$\frac{df(x)}{dx}-f(x/2)=0$$
so plugging in we get:
$$\frac{1}{2^t\ln(t)}\frac{df(2^t)}{dt}-f(2^{t-1})=0$$
or we could write it as:
$$\frac{df(2^t)}{f(2^{t-1})}=2^t\ln(t)dt$$
$$\int\frac{df(2^t)}{f(2^{t-1})}=2^t+C$$
I would assume the best way to approach this would be to assume that:
$$f(2^{t-1})=f(2^t)g(2^t)$$
or something like this
