Show that the trajectory of the solution $ x(t),y(t))$ is contained in the curve $E_{0}$. (Non-Linear Systems, ODE) I've been visiting this site for years to solve doubts, and this is my first question it's about something that's almost totally new to me.
We have the autonomous system $x'=y$ and $y'=x^3-x$ (pendulum).
And the following question:
Let $(x(t),y(t))$ be a $(C^1)$ class) solution of the nonlinear system, with $t=0$ given by $(xo,yo)$. Show that the trajectory $(x(t),y(t))$ must be contained on the curve:
$E_{0}=\dfrac{1}{2}y^{2}-\dfrac{1}{4}(x^{2}-1)^{2}, E_{0}$ its constant.
I'm very confused about this exercise, because i don't know what to do with the curve $E_0$ if it's constant!.
 A: If we set
$E(x, y) = \dfrac{1}{2}y^2 - \dfrac{1}{4}(x^2 - 1)^2, \tag 1$
we can then express $E(x, y)$ along any (sufficiently smooth, say $C^1$) curve
$\gamma(t) = (x(t), y(t)) \tag 2$
as
$E(\gamma(t)) = E(x(t)), y(t)) = \dfrac{1}{2}y^2(t) - \dfrac{1}{4}(x^2(t) - 1)^2; \tag 3$
along such a curve we may compute $dE/dt$:
$\dfrac{dE(\gamma(t))}{dt} = \dfrac{dE(x(t), y(t))}{dt} = \dfrac{\partial E(x(t), y(t))}{\partial x} x'(t) +  \dfrac{\partial E(x(t), y(t))}{\partial y} y'(t); \tag 4$
we have
$\dfrac{\partial E(x, y)}{\partial x} = -\dfrac{1}{2}(x^2 - 1)(2x) = -x(x^2 - 1), \tag 5$
and 
$\dfrac{\partial E(x, y)}{\partial y} = y; \tag 6$
if we assume that
$\gamma'(t) = (x'(t), y'(t)) = (y, x^3 - x), \tag 7$
we see that
$\dfrac{dE(\gamma(t))}{dt} =\dfrac{\partial E(x(t), y(t))}{\partial x} x'(t) +  \dfrac{\partial E(x(t), y(t))}{\partial y} y'(t)$
$= -x(t)(x^2(t) - 1)y(t) + y(t)(x^3(t) - x(t))$
$= -x^3(t)y(t) + x(t)y(t) + y(t)x^3(t) - y(t)x(t) = 0; \tag 8$
thus $E(t) = E_0$ is constant along the curves $\gamma(t)$, and from this we furthur conclude that $\gamma(t) = (x(t), y(t))$ is contained in the curve
$\dfrac{1}{2}y^2 - \dfrac{1}{4}(x^2 - 1)^2 = E_0. \tag 9$
