Who can solve this ordinary differential equation? $$f'=f(1-x)$$ This equation appears when I try to solve the eigenvalue problem of an integral equation.
 A: If, as Kaster suggested, you mean $f^\prime(x) = f(1-x)$, for all $x$, we have the following:
$$f^{\prime\prime}(x) = \frac{\mathrm{d}}{\mathrm{d}x}f(1-x) = -f^\prime(1-x) = -f(x)$$
This equation is solved by $f(x)=A\sin(x+\phi)$ and $f^\prime(x) = A\cos(x+\phi) = A\sin(\frac{\pi}{2}+x+\phi)$. Again requiring $f^\prime(x) = f(1-x)$ gives us
$A\sin(\frac{\pi}{2}+x+\phi)=A\sin(1-x+\phi)$, for all $x$.
$\sin(1-x+\phi) = \sin(\pi-1+x-\phi)$, thus if we disregard the case where $A=0$ we require
$$\sin(\frac{\pi}{2}+x+\phi)=\sin(\pi-1+x-\phi),$$
which is true for all $x$ if $\frac{\pi}{2}+\phi-\pi+1+\phi = 2\phi+1-\frac{\pi}{2}\in2\pi\mathbb{Z}$.
Thus your equation is solved by $f(x) = A\sin(x+\phi)$, where $\phi=\frac{\pi}{4}-\frac{1}{2}+k\pi$ for some $k\in\mathbb{Z}$ and where $A$ is any real number.
Since translating $\sin$ by $\pi$ is merely multiplication by $-1$, this possibility is actually already accounted for by $A$.
Hence, the only solution to your equation would be
$$f(x) = A\sin\left(x+\frac{\pi-2}{4}\right)$$
A: If you have $\frac{\rm{d} y}{\rm{d} x} = y\,(1-x)$ then you separate the variables to
$$ \frac{1}{y}\, \frac{ \rm{d} y}{\rm{d} x} =(1-x) $$
  $$ \int \frac{1}{y}\, \rm{d} y = \int (1-x) \,\rm{d}x + K $$
$$ y(x) = \boldsymbol{e}^{K}\,\boldsymbol{e}^{x-\frac{x^2}{2}} $$
