Prove that a solution to differential equation is unbounded Consider the following system:
$$ \begin{cases}
x'=x-6y\\
y'=-2x-y
\end{cases} $$
I want to prove that the solution to the system which satisfies $$ x\left(0\right)=1,\thinspace\thinspace y\left(0\right)=0 $$
is unbounded, Without solving the equation.
My work so far:
Notice that this system is hamiltonian system, meaning: $$ \begin{cases}
x'=\frac{\partial}{\partial y}H\left(x,y\right)\\
y'=-\frac{\partial}{\partial x}H\left(x,y\right)
\end{cases} $$
Where $ H\left(x,y\right)=x^{2}+xy-3y^{2} $ and the solution $\varphi$ which satisfies $\varphi(0)=(1,0)$ also satisfies $$ \varphi\left(0\right)\in\Lambda_{1}=\left\{ \left(x,y\right)\thinspace:\thinspace H\left(x,y\right)=1\right\} =\left\{ \left(x,y\right)\thinspace:\thinspace x^{2}+xy-3y^{2}=1\right\}  $$
And the level curve $\Lambda_1$ is hyperbola (which means it is unbounded).
Now, I tried to assume by contradiction that $\varphi(t) $ indeed is a bounded solution, which means that both $x(t)$ and $y(t)$ are bounded, but I cannot find how to reach a contradiction.
Any help would be appreciated.
 A: Now let me write a detailed answer using the idea in my early comments, at which time I was typing on my not-so-smart phone.
We argue by contradiction. Suppose that $x(t), y(t)$ are bounded on their maximal interval of existence, then they must exist on the whole interval $[0,\infty)$, by the standard continuation theorem (e.g. Theorem $4$ in this pdf.) Define
$$X(t)=e^{-t}x(t), Y(t)=e^{-t}y(t),\qquad t\in[0,\infty).$$
Then $X, Y$ satisfy the equations
$$ \begin{cases}
X'=-6Y,\\
Y'=-2X-2Y,
\end{cases} $$
with $X(0)=1, Y(0)=0$. It follows from the boundedness of $x(t)$ and $y(t)$ that
$$\lim_{t\to\infty}X(t)=\lim_{t\to\infty}Y(t)=0.\tag{$*$}$$
Let $f(t)= (X(t))^2-3(Y(t))^2$, then $f'(t)= 12(Y(t))^2\ge 0$ for all $t>0$, thus $f$ is non-decreasing on $[0,\infty)$. Since $f(0)=1$ by the initial data, we must have
$$f(t)\geq 1,\qquad t\in[0,\infty).$$
But $(*)$ implies $\lim_{t\to\infty}f(t)=0$. This is a contradiction!
A: It is easy to see
$$ \frac12(x^2-3y^2)'=x(x-6y)-y(-2x-3y)=x^2+3y^2\ge x^2-3y^2 $$
which implies
$$ (e^{-2t}(x^2-3y^2))'\ge0. $$
So
$$ x^2-3y^2\ge (x^2(0)+y^2(0))e^{2t}=e^{2t}. $$
This gives
$$ x^2+y^2\ge x^2-3y^2\ge e^{2t}. $$
So $(x,y)$ is unbounded.
