November
6, 2006
Find
all real functions $f$ such that, for all real $x$,
$f(x+2)=f(x) \quad\text{and}\quad
f'(x)=f(x+1)-2$.
Solution. Since
$f'(x)=f(x+1)-2$, $f'$ is differentiable and
$f''(x)=f'(x+1)=f(x+2)-2=f(x)-2$.
So $f''(x)=f(x)-2$. Solving this differential equation yields
$f(x)=2+c_1 e^x+c_2 e^{-x}$, and since $f(x+2)=f(x)$, $c_1e^2=c_1$ and
$c_2e^{-2}=c_2$ so $c_1=c_2=0$. The only function that works is then
$f(x)=2$.