Find the equilibrium position of an object which satisfies the equation of motion Find the equilibrium position of an object which satisfies the equation of motion. I know how that there is a formula $$ d^{2}x/dt^{2} = - \omega^{2}x $$ but I can't see the usage of it when there are so many "things" on the right side of the equation.
$$ 4d^{2}x / dt^{2} = −x^{3} + x^{2} − x + 1 $$ 
 A: We have
$$
F(x) = m \frac{d^2 x}{dt^2} = \frac{m}{4}(-x^3 + x^2 - x + 1)
$$
which is a force only depending on position $x$ and not on velocity $\dot{x}$ or time $t$.
We have
$$
m \ddot{x} = F(x) \iff \\
m \dot{x}\ddot{x} = F(x) \, \dot{x} \iff \\
\frac{d}{dt} \left( \frac{m}{2} \dot{x}^2 \right) = F(x) \, \dot{x}
= \frac{d}{dt}\int\limits_{t_0}^tF(x(\tau)) \, \dot{x}(\tau) \, d\tau
= \frac{d}{dt}\int\limits_{x_0}^x F(\xi) \, d\xi
$$
one sets
$$
T(v) = \frac{m}{2} v^2 \\
U(x) = - \int\limits_{x_0}^x F(\xi) \, d\xi
$$
and calls them kinetic and potential energy. They fulfill
$$
T + V = E = \text{const}
$$
At points of equilibrium $x_0$ we have a vanishing force 
$$
F(x_0) = -U'(x_0) = 0
$$ 
and if we deviate from the equilibrium position, the forces must be pushing back to the equilibrium point, so we require
$$
U''(x_0) > 0
$$
From $F(x) = 0$ we have one candidate $x_0 = 1$.

We have
$$
U''(x) = (U'(x))' = (-F(x))' = -\frac{m}{4}(-3x^2 + 2 x - 1)
$$
and
$$
U''(1) = \frac{m}{2} > 0 $$
so $x_0 = 1$ is a point of equilibrium.
Here is an image

which shows $F(x)/m$ and a $U(x)/m$.
A: Recall Newton's law
$$
\frac{{\rm d}^2 x}{{\rm d}t^2} = -\frac{{\rm d}V}{{\rm d}x}
$$
In you case, it is pretty straightforward to find the potential $V$
$$
V(x) = \frac{x^4}{16} - \frac{x^3}{12} + \frac{x^2}{8} - \frac{x}{4}
$$
You see that the dominant term is $\sim x^4$, this means that the potential goes to infinite, therefore, it must have an equilibrium point somewhere, to find it just calculate the locations where 
$$
\frac{{\rm d}V}{{\rm d}x} = 0
$$
If you do the math, you'll find the only real root is $x=1$ (red point in the graph)

