Find all $C^1$ functions from the reals to the reals satisfying $f'(a-x)=f(x)$ for some $a>0$. My try : it is easy to see that $f$ is twice differentiable so by differentiations we get :
$f''(x)=-f'(a-x)=f(x)$ and so $f''(x)+f(x)=0$ and the solution must be of the form $a\cos(x)+b
\sin(x)$ for some constants $a,b$.
Now plugging this in the original equation i get some ugly equations (well, actually not that ugly, but it seems kinda strange) on $a$ that i have to solve and i am unnsure of whether this is the right thing to do or not.
Is it the right approach ? If not not, could you show me what i've done wrong ?
By the way, if anyone could post good resources to learn basic strategies for solving differential equations of order 1 and 2 at beginner's level i would be extremely grateful to them.
$f(x)=c_1e^{ix}+c_2e^{-ix}$
$ic_1e^{i(a-x)}-ic_2e^{-i(a-x)}=c_1e^{ix}+c_2e^{-ix}$
$ic_1e^{ia}-c_2=0\\c_1+ic_2e^{-ia}=0$
Thus $f(x)=ce^{ix}+ice^{ia}e^{ix}$
$f(x)=c_1e^{ix}+c_2e^{-ix}$
$ic_1e^{i(a-x)}-ic_2e^{-i(a-x)}=c_1e^{ix}+c_2e^{-ix}$
$ic_1e^{ia}-c_2=0\\c_1+ic_2e^{-ia}=0$
Thus $f(x)=ce^{ix}+ice^{ia}e^{ix}$