Solution: Let us first derive the equation of a circle passing through the origin and whose center lying on the $x$-axis (look at the figure below). The equation of the circle will be \begin{align*} (x-a)^2 + y^2 = a^2 & \implies x^2 + a^2 - 2ax + y^2 = a^2 \\ & \implies x^2 + y^2 - 2ax = 0. \end{align*} Since the center is arbitrary, we can write the equation of all such circles as \begin{equation}\label{eq:27OCt2023-1} x^2 + y^2 + 2ax = 0, \end{equation} where $a$ is an arbitrary constant.

Now to determine the differential equation, we need to eliminate the arbitrary constant $a$. Differentiating \eqref{eq:27OCt2023-1} with respect to $x$, \begin{equation}\label{eq:27Oct2023-2} 2x + 2y \frac{\mathrm{d}y}{\mathrm{d}x} + 2a = 0. \end{equation} Substituting the value of $2a$ in \eqref{eq:27Oct2023-2} using \eqref{eq:27OCt2023-1}, we get \[ 2x + 2y\frac{\mathrm{d}y}{\mathrm{d}x} + \left( - \frac{x^2 + y^2}{x} \right) = 0, \] which simplifies to \[ 2x^2 + 2xy \frac{\mathrm{d}y}{\mathrm{d}x} - x^2 - y^2 = 0 \implies 2xy \frac{\mathrm{d}y}{\mathrm{d}x} + x^2 - y^2 = 0. \] Thus, the required differential equation will be \[ 2xy \frac{\mathrm{d}y}{\mathrm{d}x} + x^2 - y^2 = 0. \]