If f'(x)=f(x+2), then f(x)=∫f(x+2)dx.
But f(x+2)=∫f(x+4)dx, so f(x)=∫(∫f(x+4)dx)dx and f(x+4)=∫f(x+6)dx, therefore:
f(x)=∫(∫f(x+4)dx)dx=∫(∫(∫f(x+6)dx)dx)dx, and so on:
f(x)=∫(∫(∫...(∫f(x+2n)dx)...)dx)dx)dx, for integer n∈[1,∞).
Also f'(0)=f(2); f'(2)=f(4); f'(4)=f(6), etc.
We can also have:
f'(x)=f(x+2), so f"(x)=f'(x+2);
But f'(x+2)=f(x+2+2)=f(x+4) so f"(x)=f(x+4), because x is just a placeholder in the function.
This implies f(n)(x)=f(x+2n) where integer n>0 and f(n) means the nth derivative.
It also implies f(n)(0)=f(2n). Again, n∈[1,∞).
More to follow in due course...