How can I solve this rather simple looking integral equation? I was working on a physics problem and I have reduced it down to a simple integral equation with two boundary conditions:
$$\int_0^{l-t}y(x, t) dx = lh$$
With the conditions:
$$y(0, t) = y(l-t, t) = h$$
I am looking for $y(x, t)$. $l$, $h$ and $t$ are positive real numbers.
Unfortunately, I don't know how to solve it. So, I need a little help there.
Also, I am curious if this same problem could be converted into a differential equation.
 A: Let the antiderivative (wrt $x$) of $y(x,t)$ be $Y(x,t)$. Then
$$Y(l-t,t)-Y(0,t)=lh$$ and deriving on $l$,
$$y(l-t,t)=h.$$
The given conditions are automatically fulfilled, and the last identity establishes an implicit relation between $l$ and $t$, which depends on the particular expression of $y$. But you can't solve for $y$ just with the available information.
A: Has this something to do with FEM?
I would do: 
at $x=0$, $$y(x, t)=h$$
at $x=\frac{1}{2} \cdot(l-t)$, $$y(x, t)=h+\frac{2ht}{l-t}$$
at $x=(l-t)$, $$y(x, t)=h$$
Is this what you are looking for?
A: There are insufficient conditions to determine a unique solution. You can, however find a particular solution. The most obvious is one that's constant in $x$
$$ y(x,t) = \frac{hl}{l-t} $$

If you want to be fancy, you can also look for a Fourier-like solution
$$ y(x,t) = h + \sum_{n=1}^{\infty} c_n(t) \sin\left( \frac{n\pi}{l-t}x \right) $$
Then
$$ \int_0^{l-t} y(x,t)\ dx = h(l-t) + \sum_{k=0}^{\infty}c_n(t)\frac{l-t}{n\pi}(1-(-1)^n) = lh $$
$$ \frac{2(l-t)}{\pi}\sum_{k=0}^\infty \frac{c_{2k+1}(t)}{2k+1} = th $$
which gives a family of solutions
$$ y_n(x,t) = h + \frac{n\pi th}{2(l-t)}\sin\left(\frac{n\pi}{l-t}x\right), \quad n = 1,3,5, \dots $$
