Showing instablity of differential equation. Assume differential equation

$$ x'=2x+y+x \cos t-y \sin t $$
  $$ y'=-x+2y-x\cos t+y \sin t $$



*

*Show that solution $(x(t),y(t))=(0,0)$ is unstable.

*Is there a non-trival solution such that $\lim_{t\to \infty}(x(t),y(t))=(0,0)$.


for question 2 , I think there exist a non-trival solution such that $\lim_{t\to \infty}(x(t),y(t))=(0,0)$ and it shows that having zero limit at infinity doesn't show stablity.
 A: First question
Multiply the first equation by $x'$, the second by $y'$ and add:
$$
x\,x'+y\,y'=2(x^2+y^2)+x^2\cos t+y^2\,\sin t-2\,x\,y\cos t\sin t.
$$
If $0\le t\le\pi/2$ then $0\le\cos t\le1$, $0\le\sin t\le1$ and
$$\begin{align}
x^2\cos t+y^2\,\sin t-2\,x\,y\cos t\sin t&\ge x^2\cos^2 t+y^2\,\sin^2 t-2\,x\,y\cos t\sin t\\
&=(x\cos t-y\sin t)^2\\
&\ge0.
\end{align}$$
Thus, if $0\le t\le\pi/2$ then
$$
(x^2+y^2)'\ge4(x^2+y^2).
$$
Integrating
$$
x^2+y^2\ge(x(0)^2+y(0)^2)e^{4t},\quad 0\le t\le\frac\pi2.
$$
A: The equation is
$$
\begin{bmatrix}
x'\\y'
\end{bmatrix}
=
\begin{bmatrix}
2+\cos(t)&1-\sin(t)\\
-1-\cos(t)&2+\sin(t)
\end{bmatrix}
\begin{bmatrix}
x\\y
\end{bmatrix}\tag{1}
$$
Although the trace of the square matrix above is $4+\sqrt2\sin(t+\pi/4)$ which is positive for all $t$, Artem has pointed out an article in which an example is given where some of the eigenvalues have positive real part yet the system is stable.

Another approach
In an approach similar to Julián Aguirre's answer, we can write
$$
\begin{bmatrix}
x&y
\end{bmatrix}
\begin{bmatrix}
x'\\y'
\end{bmatrix}
=
\begin{bmatrix}
x&y
\end{bmatrix}
\begin{bmatrix}
2+\cos(t)&1-\sin(t)\\
-1-\cos(t)&2+\sin(t)
\end{bmatrix}
\begin{bmatrix}
x\\y
\end{bmatrix}\tag{2}
$$
Noting that the eigenvalues of the symmetric part of the matrix in $(2)$
$$
\begin{bmatrix}
2+\cos(t)&-\tfrac12(\sin(t)+\cos(t))\\
-\tfrac12(\sin(t)+\cos(t))&2+\sin(t)
\end{bmatrix}\tag{3}
$$
are
$$
\frac{4\pm\sqrt2+\sin(t)+\cos(t)}2\tag{4}
$$
we get that for all $t$,
$$
(2-\sqrt2)
\begin{bmatrix}
x&y
\end{bmatrix}
\begin{bmatrix}
x\\y
\end{bmatrix}
\le\begin{bmatrix}
x&y
\end{bmatrix}
\begin{bmatrix}
x'\\y'
\end{bmatrix}
\le(2+\sqrt2)
\begin{bmatrix}
x&y
\end{bmatrix}
\begin{bmatrix}
x\\y
\end{bmatrix}\tag{5}
$$
That is,
$$
(4-2\sqrt2)(x^2+y^2)\le(x^2+y^2)'\le(4+2\sqrt2)(x^2+y^2)\tag{6}
$$
which implies that
$$
e^{(4-2\sqrt2)t}(x_0^2+y_0^2)\le x^2+y^2\le e^{(4+2\sqrt2)t}(x_0^2+y_0^2)\tag{7}
$$
Inequality $(7)$ shows that $(0,0)$ is not stable and is the only solution for which $\lim\limits_{t\to\infty}(x,y)=(0,0)$ (in fact, the only solution for which $(x,y)$ remains bounded).
A: I do not know which stability criterion should be applied, so I try a simple argument:
An object at the origin $(0,0)$ which is disturbed by a small positive displacement $\epsilon$ in $x$-direction will experience a velocity $x' = (2 + \cos t) \epsilon > 0$, it will move away from the origin.
For a small disturbance $\epsilon$ in $y$-direction it will experience a velocity $y' = (2 + \sin t) \epsilon > 0$, which will move the object away as well.
About the second part I am not sure, because of the instability at $(0,0)$, something there could only stay, if undisturbed.
Here is a plot of the velocity field for $t=0$:

