transform integral to differential equations I found a similar system of integral equations in a paper. It says that it can be solved by differentiating and then using standard techniques. My question is, how can I differentiate such a system in general? Do I have to differentiate for $x$ or $y$ or for both?
$$f(y)+f(x) + xh(x) + \int_x^y zg(z)dz=a, \qquad \forall x <y $$
$$f(x)+h(x) + \int_x^\infty g(z)dz=1 $$
$$f(x), h(x),\qquad g(z) \geq 0$$
where $x$ and $y$ are variables in $\mathbb{R}$ and $f,g,h$ are functions to be determined. 
I would be very glad for any help on how to proceed here!
 A: You have three unknowns so in order to solve the system, you'll need (at least) three equations. If we differentiate the first with respect to $y$ (and replacing $y$ with $x$), we get
$$f'(x) + xg(x) = 0. \tag{1}$$
If we differentiate the first equation instead with respect to $x$, we get
$$f'(x) + h(x) + xh'(x) - xg(x) = 0. \tag{2}$$
Differentiating the second equation with respect to $x$, we get
$$f'(x) + h'(x) - g(x) = 0. \tag{3}$$
Now we have three equations and some hope to solve the system. Adding $(1)$ and $(2)$ yields
$$2f'(x) + h(x) + xh'(x) = 0 \Longrightarrow \frac{d}{dx}\left(2f(x)+ xh(x)\right) = 0.$$
Hence $2f(x) + xh(x) = C$, i.e. $h(x) = \frac{C-2f(x)}{x}$. From $(1)$, $g(x) = -\frac{1}{x}f'(x)$. But from $(3)$, we also have that
$$g(x) = (f(x)+h(x))' = \frac{x(f(x)+xf'(x)-2f'(x))-(C+xf(x)-2f(x))}{x^2}$$
which simplifies to
$$g(x) = \frac{x^2f'(x)-2xf'(x)+ 2f(x)-C}{x^2}.$$
Setting the two expressions for $g$ equal to each other gives
$$-\frac{1}{x}f'(x) = \frac{x^2f'(x)-2xf'(x)+ 2f(x)-C}{x^2},$$
i.e.
$$-xf'(x) = x^2f'(x) -2xf'(x)+ 2f(x)-C.$$
So $f$ solves the differential equation
$$(x^2-x)f'(x) + 2f(x)-C = 0.$$
This can be solved by integrating factors. From this you can get $g$ and $h$. To get the constants of integration, you'll likely need to use your original equations. Of course you'll have to check that each of $f$, $g$ and $h$ are non-negative functions.
