I would like to solve this differential equation that is similar to the delayed differential equation here.
this is the DE: $f'(x) = f(kx), \;$ for some real $k$.
The reason for attempting this problem is that I am interested in solving this DE but do not have the means to solve it:
$øf(øx) = 2f(x)∫f(t)\mathrm{d}t$ integrated over $[0,x]$.
I though I'd start with a simpler yet similar problem to help me understand the original question.
If you write formally $f(x) = \sum_{i=0}^\infty a_i x^i$, your equation $f'(x) = f(kx)$ becomes $$ \sum_{i=0}^\infty ((i+1) a_{i+1} - k^i a_i) x^i$$ The solution to the recurrence $(i+1) a_{i+1} = k^i a_i$ is $$ a_i = a_0 \frac{k^{i(i-1)/2}}{i!}$$ The series $$ \sum_{i=0}^\infty \frac{k^{i(i-1)/2}}{i!} x^i$$ converges to an entire function of $x$ for $|k| < 1$.