Let $ f$ be a periodic function of period $T$ and $\phi $ an solution of the linear equation
Prove that if $ \phi(x)$ is a solution, then $\phi(x+T) $ is also a solution. More than this, prove that there exists a constant $c$ such that
$$\phi(x+T) = c \phi(x)$$
Solving the linear ode, we have that $$y=k \exp(\int - f(x) dx)$$
If $\phi$ is a solution, then $\phi(x) =k \exp(\int - f(x) dx)$
I dont know how to rigorously proceed from here: We know that f(x)=f(x+T) since the function is periodic. So, is it sufficient to substitute f(x+T) for f(x) when calculating \phi(x+T)? I am not sure how I should work with periodic functions and indefinitive integrals. Maybe it is more convenient tp write
$$y= k \exp(\int -f(x)dx + C_1)$$
and when calculating $\phi(x+T)$, this constant $C_1$ changes for another constant, we say $C_2.$. But zgain, I am not sure if it is correct.