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Find all functions $f(x)$ satisfying $f(x)+f^{\prime}(\pi-x)=1$ for all $x \in \mathbb{R}$.

This is a question from Moscow. I have tried $f(x)=x^m$ and it clearly does not work. Clearly $f(x)=1$ works. But I have no idea how to obtain remaining functions which satisfy the equation.

Clearly $f^{\prime}(0)=1-f(\pi)$.

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  • $\begingroup$ Just a random comment, I think this is a good hint. Let's say that $f(x)=1+h(x)$ satisfy the functional equation. Therefore $1+h(x)+h'(\pi-x)=1$, hence $h(x)=-h'(\pi-x)$. $\endgroup$
    – rafforaffo
    May 20, 2015 at 9:27

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Notice that the original equation we are given tells us that $f$ and $f'$ both belong to the same class of differentiable functions. This means we can restrict ourselves to $C^\infty$ functions, and it makes sense to take derivatives. Hence $$ f'(x) - f''(\pi-x) = 0.$$ The original equation also gives a relation between $f$ and $f'$. By changing variables to $y = \pi-x$ we can substitute to get $$f''(y) +f(y) =1$$ Which is a second order, non-homogeneous ODE, and thus has a unique solution; ie the one proposed intuitively by Martigan.

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Intuitively you think about trigonometric functions.

Let's try $f(x)=\alpha \cos x +\beta \sin x +\gamma$

$f'(x)=-\alpha \sin x +\beta \cos x$

$f'(\pi-x)=-\alpha \sin x -\beta \cos x$

$f(x)+f'(\pi-x)=(\alpha -\beta)(\cos x - \sin x)+\gamma=1$

Hence $\alpha- \beta=0$ and $\gamma=1$.

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  • $\begingroup$ Now we have a solution. But what about all solutions? $\endgroup$
    – gebruiker
    May 20, 2015 at 9:57

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