I take the next exercise from Apostol's book. I thought for a while , but nothing. Then I read the solution. I try to think: how did he get it , but nothing. I want to know how to get to the solution. So ...
exercise : A periodic function with period $a$ satisfies $f(x + a) =f(x)$ for a11 $x$ in its domain. What can you conclude about a function which has a derivative everywhere and satisfies an equation of the form $$f(x+a) = bf(x)$$ for a11 x, where $a$ and $b$ are positive constants?
solution: $f(x) = b^{\frac{x}{a}}g(x)$, where $g$ is periodic with period $a$.