The differential equation of all circles passing through the origin and having their centres on the $x$ -axis is:

A circle with centre on the $x$ -axis has coordinates $(a,0)$.If it passes through the origin $(0,0)$, the radius must be $|a|$.The equation of the circle is $(x-a{)}^2+y^2=a^2$.
$$x^2-2ax+a^2+y^2=a^2$$
$$x^2+y^2=2ax\dots (1)$$
Differentiating with respect to $x$:
$$2x+2y\frac{dy}{dx}=2a\dots (2)$$
Substitute $2a$ from (2) into (1):
$$x^2+y^2=\left(2x+2y\frac{dy}{dx}\right)x$$
$$x^2+y^2=2x^2+2xy\frac{dy}{dx}$$
$$y^2=x^2+2xy\frac{dy}{dx}$$