Can the following functional equation be reduced to a standard ODE? Consider the functional equation
$$
f\left(x\right) := \dfrac{x^3}{2}\left\{G''\left(x\right) k + \int_{\underline{x}}^{x}\dfrac{\partial^2D_H\left(x,y\right)}{\partial x^2}G'\left(y\right)d y + \int_{x}^{\overline{x}}\dfrac{\partial^2D_L\left(x,y\right)}{\partial x^2}G'\left(y\right)d y\right\} = \operatorname{const}
$$
for $x \in [\underline{x}, \overline{x}]$.
The functions $D_H\left(x,y\right)$ and $D_L\left(x,y\right)$ are known and well-behaved, with the property that $D_H\left(x,x\right) - D_L\left(x,x\right) = k$ is a (negative) constant.
Ultimately, I'm looking for a solution $G\left(x\right)$ with the following properties:
(1) $G\left(\underline{x}\right) = 1 - G\left(\overline{x}\right) = 0$
(2) $G'\left(x\right) \geq 0$ for all $x \in [\underline{x}, \overline{x}]$
(3) $G'\left(\overline{x}\right) = 0$.
It appears that this is in general a difficult problem. However, I'm wondering whether at least some numerical solution candidates can be obtained.
For this, it would be great if the equation could be reduced to a standard ordinary differential equation. For example, with
\begin{align}
D_H\left(x,y\right) &= \dfrac{\left(1-x\right)^2}{2} - t + \left(1-x\right)y
&\text{ and }&&
D_L\left(x,y\right) &= \dfrac{1}{2} + t - x + y + \dfrac{y^2}{2},
\end{align}
which are obtained through a simple example in the underlying model, the functional equation becomes
$$
f\left(x\right) = \dfrac{x^3}{2}\big\{-2t G''\left(x\right) + G\left(x\right)\big\} = \operatorname{const},
$$
which can be solved semi-analytically.
So I'm wondering whether the functional equation can be reduced to a standard ODE under fairly general conditions, or whether I'm out of luck. Any ideas, even regarding sources where to look up similar problems, would be appreciated.
 A: At first,
\begin{align}
&\dfrac{d}{dx} \left(\int\limits_{\underline{x}}^{x}D_H(x,y)G'(y)\,dy\right) = D_H(x,x)G'(x) + \int\limits_{\underline{x}}^{x}\dfrac{\partial D_H(x,y)}{\partial x}G'(y)\,dy,\\
&\dfrac{d}{dx} \left(\int\limits_{x}^{\overline x}D_L(x,y)G'(y)\,dy\right) = -D_L(x,x)G'(x) + \int\limits_{x}^{\overline x}\dfrac{\partial D_L(x,y)}{\partial x}G'(y)\,dy,\\
&\dfrac{d}{dx} \left(\int\limits_{\underline{x}}^{x}D_HG'\left(y\right)\,dy + \int\limits_{x}^{\overline{x}}D_LG'\left(y\right)\,dy\right) =\\
&\qquad{kG'(x) + \int\limits_{\underline x}^{x}\dfrac{\partial D_H(x,y)}{\partial x}G'(y)\,dy + \int\limits_{x}^{\overline x}\dfrac{\partial D_L(x,y)}{\partial x}G'(y)\,dy},\\
&\dfrac{d^2}{dx^2} \left(\int\limits_{\underline{x}}^{x}D_HG'\left(y\right)\,dy + \int\limits_{x}^{\overline{x}}D_LG'\left(y\right)\,dy\right) = kG''(x) + \dfrac{\partial}{\partial x}(D_H(x,x)-D_L(x,x))\cdot G'(x) +\\
&\qquad \int\limits_{x}^{\overline x}\dfrac{\partial^2D_H(x,y)}{\partial x}G'(y)\,dy + \int\limits_{x}^{\overline x}\dfrac{\partial^2D_L(x,y)}{\partial x}G'(y)\,dy,\\
&\dfrac{x^3}2\dfrac{d^2}{dx^2}
\left(\int\limits_{\underline{x}}^{x}D_H(x,y)G'\left(y\right)d y + \int\limits_{x}^{\overline{x}}D_L(x,y)G'\left(y\right)d y\right) = const,\\
&\int\limits_{\underline{x}}^{x}D_H(x,y)G'(y)\,dy + \int\limits_{x}^{\overline{x}}D_L(x,y)G'(y)dy = \dfrac{C_{-1}}x+C_0+C_1x.\tag1\\
\end{align}
By parts:
\begin{align}
&D_H(x,y)G(y)\Big|_{y=x}^{y=\overline x}+D_L(x,y)G(y)\Big|_{y=\underline x}^{y=x}+\int\limits_{\underline{x}}^{x}\dfrac{\partial D_H(x,y)}{\partial y}G(y)dy + \int\limits_{x}^{\overline{x}}\dfrac{\partial D_L(x,y)}{\partial y}G(y)dy = \dfrac{C_{-1}}x+C_0+C_1x,\\
&D_H(x,\overline x) - kG(x) + \int\limits_{\underline{x}}^{x}\dfrac{\partial D_H(x,y)}{\partial y}G(y)dy + \int\limits_{x}^{\overline{x}}\dfrac{\partial D_L(x,y)}{\partial y}G(y)dy = \dfrac{C_{-1}}x+C_0+C_1x.\tag2\\
\end{align}
The equation in the forms $(1), (2)$ allows to use various solution methods, i.e. undefined coefficients method for the known form of solution or its approximation with series.
Also there is a possibility to obtain linear system for values $G(y)$, using Simpson's approximation for integrals in $(2).$
