A very natural and interesting question.
If the paper stays nearly flat, we may use the linearized approximation. I will neglect the gravitational potential energy because it's probably much smaller than the bending energy (note that the shape of the bent paper is almost the same when the desk is vertical as when it is horizontal).
The paper doesn't want to bend much, so the energy contains a term that punishes the second derivatives of $y$ (the curvature)
$$E = K \int_0^L {\rm d}x\,(y'')^2$$
where $y'\equiv dy/dx$, and so on. We want to keep the length of the paper fixed - it's prescribed by the way how you attach it at the end points. In the linear approximation, the length is a linear function of
$$D = \int_0^L {\rm d} x\, (y')^2 $$
which may be seen, if you need it, by a Taylor expansion of the exact expression for the length of the graph, $\int (1+y^{\prime 2})^{1/2}{\rm d}x$. Add it with a Lagrange multiplier, requiring
$$\delta (E+\lambda D) = 0.$$
I think that the resulting equations say
$$ y'''' = \frac{\lambda}{K} y''.$$
Note that in the Euler-Lagrange equations, all the primes "clump". If you write $Y=y''$, it says that the second derivative of $Y$ is proportional to $Y$ itself. The physically relevant solution is
$$ Y = Y_0 \cos (2\pi x n / L) $$
where only $n=1$ is possible if there is a table underneath the paper. Consequently,
$$ y = y_0 [1-\cos (2\pi x / L)],$$
too. I added the term $1$ as an integration constant (while the other is zero) to guarantee that $y=0$ and $y'=0$ at the boundaries.
So what you get in practice is one wave of a cosine, from one minimum to the next one.
A more detailed discussion would be needed to eliminate the other cosine-like function with the same frequency, the sine, as well as linear functions, and to discuss whether the exponentially increasing/decreasing solutions may ever be relevant. Well, it wouldn't be too difficult. A more general solution for the same (cosine-like) sign of $\lambda$ would be
$$ y = y_0 [1-\cos (\phi_0+ 2\pi x N / L)] + Ax^2+B,$$
and the four conditions $y=0$, $y'=0$ at $x=0,L$ as well as the fixed length of the paper would imply $N=1$, $\phi_0=0$, $A=0$, $B=0$, as well as the right normalization $y_0$. As suggested above, the conditions would have other solutions where $N=2,3,4\dots$ but these solutions wouldn't satisfy $y\geq 0$ for $0\leq x \leq L$, so the paper couldn't arrange itself above the table. If there were a hole in the table, these higher harmonics (several periods of the wave containing) solutions would be possible - but I guess that they would be unstable.
I am not sure whether I would be able to solve it analytically if the angle $y'$ were not infinitesimal. It seems clear that the exact solution for significant angles is not just a cosine: the bent paper tends to resemble a balloon - for which $y(x)$ is not even single-valued - when there's too much redundant paper in the middle.
On the other hand, if you sharply bend the paper at $x=0,L$ so that $y'$ may be arbitrary at these two points, the paper in between will be bent as an arc of a circle - see another answer below - and this result is accurate even if $y'$ is of order one. Note that in the linearized approximation, the arc gives $y$ being a quadratic function of $x$ so that $y''=0$ still solves the fourth-order equation above.
