Show that every solution of the system $$ \frac{dx}{dt}=y,\quad\frac{dy}{dt}=x-x^7,\qquad(x,y)\in\mathbb{R}^2, $$ is defined for all $t\in (-\infty,+\infty)$.
The given solution is the following:
The Hamiltonian $H(x,y)=\frac{y^2}{2}-\frac{x^2}{2}+\frac{x^8}{8}$ is conserved: $$ \frac{dH}{dt}=(x^7-x)y+y(x-x^7)=0. $$ Therefore $x(t)$ and $y(t)$ remain bounded for all $t$ (otherwise $H(x,y)$ would grow). Hence $(x(t),y(t))$ is globally defined.
I cannot fully understand this argumentation...
Of course, if $(x(t),y(t))$ is a solution, the Hamiltonian is constant along $(x(t),y(t))$.
But does this tell anything about boundedness of the solution?
In fact, if $H$ would grow, i.e. $\frac{dH}{dt}>0$, then $(x(t),y(t))$ would be no solution, since the Hamiltonian is constant along solutions.
Take any particular solution, with initial conditions $x(a)=b,y(a)=c$. Then $H(x,y)=\frac{c^2}2-\frac{b^2}2+\frac{b^8}8=d$ for all $t$. If $x(t)^2$ ever got above $2d+3/4$, there would be no possible value for $y(t)$.
So $x(t)$ is bounded. $y(t)$ is also bounded, for similar reasons.
Both $x(t)$ and $y(t)$ have bounded derivative, so they can be extended into the future.