Let $g(x),h(x), f’(x) $ be functions that can be expressed by radicals , log and exp , but $f(x) $ can not.
Now consider functional equations like
$$ f(x) + c \space f(g(x)) = h(x) $$
Where $c^2 = 1$
The only solutions with h(x) not identically 0 I know are based on $g(x) = a + b x $ with solutions like $li(x)$.
How to find more solutions ?
Is $ f(x) $ always of the form $ \int y(g(x)) t(x) dx $ ?
In particular I wonder about it when iterations of $ g (x) $ are cyclic.
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$ f(x) + f(-x) = 0 $ is not intresting ofcourse.
Oh, I don't know... your last eqn might actually be pointing the way...
In any case, if you want "easy instances", consider $g(x)=a/x$, so $$ f(x)\pm f(a/x)= h(x). $$ It is evident that $$ h(a/x)=\pm h(x). $$ You then see that $$ f(x)=\frac{1}{2} h(x) +k(x,a/x), $$ where k is an antisymmetric (symmetric, respectively) function of its two arguments, solving the homogeneous equation. (Indeed, exponentiating the arguments of your trivial one gets one there...)
Similarly, for $g(x)=\sqrt{a^2-x^2}$, $$ f(x)\pm f\left (\sqrt{a^2-x^2}\right ) = h(x), \qquad x\in [0,a], \quad h(x)=\pm h\left ( \sqrt{a^2-x^2}\right ) . $$ You then see that $$ f(x)=\frac{1}{2} h(x) +k\left ( x,\sqrt{a^2-x^2}\right ) , $$ where k is an antisymmetric (symmetric, respectively) function of its two arguments, solving the homogeneous equation.
Likewise, cycling around Babbage's equation solution, observe that $$ f(x) + f\left ( \frac{a-x}{1+bx} \right)=0, $$ is solved by $$ k \left ( x, \frac{a-x}{1+bx} \right), $$ provided k be antisymmetric in its two arguments... and so on. I am only a physicist, so I don't have the complete beautiful machinery of Babbage's equation orbits under my belt...
So, the silly takeaway is to "stretch" the trivial cases by less trivial-looking transformations, conjugacies, etc... This was, in fact, the strategy Ernst Schroeder used 150 years ago on his eponymous equation solutions...