Prove that, every solution of the scalar system: $\dfrac{dx}{dt}=y,\dfrac{dy}{dt}=-\dfrac{2y}{t},(t \ge 1) $ is bounded in domain $[1, +\infty)$ EDITED
I have a problem:
Prove that, every solution of the scalar system:
$$\dfrac{dx}{dt}=y,\dfrac{dy}{dt}=-\dfrac{2y}{t},(t \ge 1) $$ is bounded in domain $[1, +\infty)$, but this system's not uniformly stable in domain $1 \le t, t_0 < +\infty$.
I have thought about my problem. But I still have no solution. 
Any help will be appreciated. Thanks!
 A: Hints:
We can solve this system and obtain:


*

*$x(t) = -\dfrac{c_1}{t} + c_2$

*$y(t) = \dfrac{c_1}{t^2}$

*We can verify that these do indeed solve the system.

*We now have to show that these solutions are bounded or unbounded (use the definition and find the lower and upper bound of each over the defined range).


What conclusions can we draw about the stability of the system?
A: $1/$ We'll show that $x,y$ are bounded in $\left [1, \infty  \right )$;


*

*Since $$\dfrac{dy}{dt}=-\dfrac{2y}{t} \implies \dfrac{dy}{y}=-2\dfrac{dt}{t}\implies \int\dfrac{dy}{y}=\int-2\dfrac{dt}{t}\implies \ln|y|=\ln\frac{1}{t^2}+\ln|c_1|$$ Therefore $y=\dfrac{c_1}{t^2}$.

*Since $\dfrac{dx}{dt}=y\implies x=\int\dfrac{c_1\mathrm{d}t}{t^2}=-\dfrac{c_1}{t}+c_2, \forall c_1,c_2 \in \mathbb{R}$.
Thus, we have $$\begin{cases}
 & \mathrm{  } x= -\dfrac{c_1}{t}+c_2\\ 
 & \mathrm{  } y= \dfrac{c_1}{t^2}.
\end{cases}$$
$|x(t)|=\left |-\dfrac{c_1}{t}+c_2  \right |\le \left |\dfrac{c_1}{t}  \right |+\left | c_2 \right | \overset{t\ge 1}{\rightarrow} |x(t)|\le |c_1|+|c_2|$. 
And $|y(t)|=\left | \dfrac{c_1}{t^2} \right |\le \left |  c_1\right |, \forall t\ge 1$.
$2/$ This system is not uniformly stable:


*

*We'll find $X(t)$ is the fundamental matrix of the scalar system:


We have $(x(t),y(t))=(-\dfrac{c_1}{t}+c_2;\dfrac{c_1}{t^2})=c_1(-\dfrac{1}{t},\dfrac{1}{t^2})+c_2(1,0)=c_1u+c_2v$. 
Thus, $u,v$ are solutions of the scalar system (with $c_1=1,c_2=0$ and $c_1=0,c_2=1$).
Moreover, if $\alpha u+\beta v=0$ then
$$\begin{cases}
 & \mathrm{  } -\dfrac{\alpha}{t}+\beta= 0\\ 
 & \mathrm{  } -\dfrac{\alpha}{t^2}=0.
\end{cases}$$
$\implies \alpha=\beta=0$. So the system $\{u,v\}$ is linearly independent.
Hence, $$X(t)=\begin{pmatrix}
 & -\dfrac{1}{t}& 1\\ 
 & \dfrac{1}{t^2}& 0
\end{pmatrix}$$
is the fundamental matrix of the scalar system.
It means the Cauchy matrix: $K(t,s)=X(t)X^{-1}(s)$ is not bounded.


*

*We have $$K(t,s)=\begin{pmatrix}
 & -\dfrac{1}{t}& 1\\ 
 & \dfrac{1}{t^2}& 0
\end{pmatrix}\cdot \begin{pmatrix}
 & 0& s^2\\ 
 & 1& s
\end{pmatrix}=\begin{pmatrix}
 & 1& -\dfrac{s^2}{t}+s\\ 
 & 0& \dfrac{s^2}{t^2}
\end{pmatrix}$$

*Now, we assume that $K(t,s)$ is bounded, then $\exists M>0$ such that:
$$\|K(t,s)\| \le M, \forall t \ge s \ge 1$$
Taking $t=s^2\ge s\ge 1 \implies -\dfrac{s^2}{t}+s=-1+s \to +\infty$, when $s \to \infty (\text{conflict}). \blacksquare $.
QED.
