Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives, such that $F(x-1)+f(1-x)=x^3$ Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives $F:\mathbb{R}\to\mathbb{R}$, such that $F(x-1)+f(1-x)=x^3$ for any $x\in\mathbb{R}$. For $1-x$ instead of $x$, $F(-x)+f(x)=(1-x)^3$, $F(x)+f(-x)=(1+x)^3$ (1). 
\begin{align*}
\left(\frac{F(x)-F(-x)}{e^x}\right)' &=\frac{(f(x)+f(-x))-(F(x)+F(-x))}{e^x}\\ &=\left(\frac{F(x)}{e^x}\right)'-\left(\frac{F(-x)}{e^x}\right)'.
\end{align*}
From here on however I don't know how to solve it. I feel that the two relations from (1) or similar ones can be used.
 A: So, we're solving for differentiable $g$ such that $g(-x) + g'(x) = (1 - x)^3$. Note that $g'(x) = (1 - x)^3 - g(-x)$, and hence $g$ is twice differentiable with
$$g''(x) = -3(1 - x)^2 + g'(-x) = -3(1 - x)^2 + (1 + x)^3 - g(x)$$
That is, $g$ must satisfy the differential equation
$$y'' + y = -3(1 - x)^2 + (1 + x)^3.$$
This differential equation has the general solution
$$y = A \cos(x) + B \sin(x) + x^3 + 3x - 2.$$
But, of course, you're looking for $f = g'$, so the possible solutions we have are
$$f(x) = -A \sin(x) + B \cos(x) + 3x^2 + 3,$$
with corresponding primitive as above.
Do all such functions work? We have
\begin{align*}f(x) + F(-x) &= -A\sin(x) + B\cos(x) + 3x^2 + 3 + A\cos(-x) + B\sin(-x) + (-x)^3 + 3(-x) - 2 \\
&= (1 - x)^3 + (A + B)\cos(x) - (A + B)\sin(x).
\end{align*}
Therefore, we only get a solution with $A = -B$. Thus, the general solution is
$$f(x) = A(\sin(x) + \cos(x)) + 3x^2 + 3,$$
with corresponding primitive
$$F(x) = A(\sin(x) - \cos(x)) + x^3 + 3x - 2.$$
A: As you note, the function $f$ with primitive $F$ satisfies $F(x-1)+f(1-x)=x^3$ if and only if
$$F(x)+f(-x)=(x+1)^3.$$
The primitive $F(x)$ uniquely determines $f(x)$, its derivative, so we'll focus solely on $F(x)$. If $F(x)$ is twice differentiable, or equivalently, if $f$ is differentiable, then differentiating the functional equation
$$F(x)+F'(-x)=(x+1)^3
\qquad\text{ yields }\qquad
F(x)-F''(-x)=3(x+1)^2.$$
Plugging $-x$ into the latter and subtracting the two equations yields
$$F(x)+F''(x)=(x+1)^3-3(-x+1)^2=x^3+9x-2.$$
This is a second order linear non-homogeneous ODE, and the standard method yields
$$F(x)=c_1\cos(x)+c_2\sin(x),$$
as the solution to the homogeneous ODE. For the non-homogeneous part, the form suggests that taking $F(x)$ a cubic polynomial will yield a solution. Setting
$F(x)=ax^3+bx^2+cx+d$ yields
$$x^3+9x-2=F(x)+F''(x)=ax^3+bx^2+(6a+c)x+(2b+d),$$
which shows that $a=1$, $b=0$, $c=3$ and $d=-2$. Hence every such $F$ is of the form
$$F(x)=c_1\cos(x)+c_2\sin(x)+x^3+3x-2.$$
Plugging this back into the original equation yields
\begin{eqnarray*}
F(x)+f(-x)
&=&(c_1\cos(x)+c_2\sin(x)+x^3+3x-2)\\
&\ &+(-c_1\sin(-x)+c_2\cos(-x)+3x^2+3)\\
&=&(c_1+c_2)\cos(x)+(c_1+c_2)\sin(x)+x^3+3x^2+3x_1)\\
&=&(c_1+c_2)(\cos(x)+\sin(x))+(x+1)^3.
\end{eqnarray*}
This shows that the function equation is met if and only if $c_1=-c_2$, and so
\begin{eqnarray*}
F(x)&=&c(\cos(x)-\sin(x))+x^3+3x-2\\
f(x)&=&-c(\cos(x)+\sin(x))+3x^2+3.
\end{eqnarray*}
About nondifferentiable solutions $f$ I have no idea at all.

Edit: As Theo Bendit notes in the comments, and shows in his answer, the fact that
$$f(x)=(1-x)^3-F(-x),$$
implies that $f$ is differentiable. So the solutions above are all solutions.
