Homogeneous universe
The homogeneous universe corresponds to a universe filled with a matter of constant density at every point.
Newtonian mechanics
In Newtonian mechanics, the homogeneous universe is simply done by considering a constant mass density
\begin{equation} \Delta \phi = 4\pi G \rho \end{equation}The Green's function of the Laplace equation is
\begin{equation} G(\vec{x}, \vec{y}) = - \frac{1}{4\pi} \frac{1}{\| \vec{x} - \vec{y} \|} \end{equation}So that our solution will be
\begin{eqnarray} \phi(\vec{x}) = \int - \frac{G \rho}{\| \vec{x} - \vec{y} \|} dy \end{eqnarray}In spherical coordinates, this will be
\begin{eqnarray} \phi(r, \theta, \phi) &=& 4\pi \int - \frac{G \rho}{\sqrt{r^2 - r_y^2}} dr_y\\ &=& 4\pi G \rho \left[\ln(\sqrt{r_y^2 - r^2} + r_y) \right]_{0}^\infty\\ &=& 4\pi G \rho \left[\lim_{r_y \to \infty}\left(\ln(\sqrt{r_y^2 - r^2} + r_y)\right) - \ln(r) \right] \end{eqnarray}The integral diverges here and this is very much what should happen : it's an old problem of Newtonian gravity that it cannot be used to describe properly an infinite, static, homogeneous universe. It can be shown that, using different methods, one can find different gravitational forces depending on how the limit is taken. It's possible to find all manners of values, finite or infinite, for the gravitational field, gravitational force and tidal forces, depending on how the regularization is performed. This is due to the boundary problem of the Laplace equation,
\begin{eqnarray} \Delta f = 0 \end{eqnarray}which fails to be unique without some proper boundaries. If we are given two solutions of the Laplace equation, $f_1$ and $f_2$, then their difference should also be a solution.
\begin{eqnarray} \Delta (f_1 - f_2) = 0 \end{eqnarray}As the Laplace equation is rotationally invariant, it can be solved from a function decomposed into spherical harmonics as a basis, ie,
\begin{eqnarray} f = \sum_{l = 0}^\infty \sum_{m = -l}^l A_{l}^m r^l Y_{m}^{l}(\theta, \varphi) \end{eqnarray}If we assume zero at $r \to \infty$, then any term for $l > 0$ must vanish :
\begin{eqnarray} f = A Y_{0}^{0}(\theta, \varphi) = A \end{eqnarray}Meaning that $f = 0$. If we drop that assumption (as we must, since its derivatives never vanish even at infinity), then we could add any such function