Finding equilibrium point from a differential equation I have the differential equation of a system as given below, $$ m\frac{d^2H}{dt^2} = mg-k\frac{I^2}{H^2} $$
How do I find the equilibrium points of this?
 A: The equilibrium points of the ordinary differential equation
$m \dfrac{d^2H}{dt^2} = mg -  k \dfrac{I^2}{H^2} \tag 1$
are characterized by the conditions
$ \dfrac{dH}{dt} = 0, \tag 2$
$ \dfrac{d^2H}{dt^2} = 0;\tag 3$
substituting (3) into (1) we find that the equilibria occur where
$ mg = k \dfrac{I^2}{H^2}, \tag 4$
that is, where
$H^2 = \dfrac{kI^2}{mg}; \tag 5$
thus, at equilibria
$H = \pm \sqrt{\dfrac{kI^2}{mg}}. \tag 6$
Note Added in Edit, Friday 10 September 2021 8:32 PM PST:  Tbe reader might well be curious as to why, when the equilibria of (1) are characterized by conditions (2), (3), we only need find the values of $H$ corresponding to (3).  The answer may be found in the structure of the phase plane for the differential equation (1).  A second-order equation such as (1) may be thought of as a case of the general second-order system
$\ddot x = f(x, \dot x), \tag 7$
which in turn may be thought as a first-order system in two variables by taking
$y = \dot x, \tag 8$
so that
$\dot y =\ddot x, \tag 9$
and hence (7) takes the form
$\dot x = y, \tag{10}$
$\dot y = f(x, y), \tag{11}$
and the equilibria of this system ((10)-(11)) occur at those points $(x, y)$ where
$\dot x = \dot y = 0; \tag{12}$
(10) and (12) taken in concert yield
$y = \dot x = 0, \tag{13}$
and now (12) and (13) allow (11) to be written
$f(x, 0) = 0, \tag{14}$
which defines and constrains the relevant values of $x$, the ones corresponding to critical points of the system of ODEs (11)-(12).  We note that in accord with (4), all these critical points lie on the $x$-axis.
The developments expressed in this note are directly applicable to equation (1) above, which may indeed be written in the form
$\dfrac{d^2H}{dt^2} = g -  k \dfrac{I^2}{mH^2} \tag{15}$
End of Note.
A: HINT
You can solve such ODE by multiplying both sides by $\dot{H}$.
