x -> 2-x

f'(2-x)=f(2-2+x)=f(x) => f''(x)=f'(2-x)=f(x) => f''(x)=f(x) => ##f(x)=c_1e^x+c_2e^{-1}##.

So, ##c_1e^x-c_2e^{-x}=c_1e^{2-x}+c_2e^{x-2}## => ##-c_2e^{-x}=c_1e^2e^{-x}## => ##-c_2=c_1e^2##. And, similarly, ##c_1=c_2e^{-2}##. So ##-c_2=c_2e^2e^{-2}## => ##c_2=c_1=0## => f(x)=0.

Is this correct? There is another solution? Thank you!