which of the following options is correct? 
Let $y(t)=\begin{pmatrix}
y_1(t)\\ 
y_2(t)
\end{pmatrix}$ satisfy  $\dfrac {dy}{dt}=Ay; t>0, y(0)=\begin{pmatrix}
0\\ 
1
\end{pmatrix}$ where $A$ is a $2 \times 2$ constant matrix with real entries satisfying trace $A=0$  and det $A >0.$ Then which of the following options are true?
  1. $y_1(t)$ and $y_2(t)$ both are oscillating functions of $t$.
  2. monotonically decreasing functions of $t$
  3. monotonically increasing functions of $t$
  4. constant functions of $t$.  

My Attempt: We take $A=\begin{pmatrix}
2 &-6 \\ 
 1& -2
\end{pmatrix}$  so that  trace $A=0$  and det $A >0.$ Now $\dfrac {d}{dt} \begin{pmatrix}
y_1(t)\\ 
y_2(t)
\end{pmatrix}=\begin{pmatrix}
2 &-6 \\ 
 1& -2
\end{pmatrix}\begin{pmatrix}
y_1(t)\\ 
y_2(t)
\end{pmatrix} \implies \begin{pmatrix}
y_1'(t)\\ 
y_2'(t)
\end{pmatrix}=\begin{pmatrix}
2y_1-6y_2\\ 
y_1-2y_2
\end{pmatrix}$. Now $y_1'(t)=2y_1-6y_2 \implies y_1''(t)=2y_1'-6y_2'=2y_1'-6(y_1-2y_2') \implies y_1''+2y_1=0.$ Hence $y(x)=c_1 \cos \sqrt 2x+c_2 \sin \sqrt 2x$ which is an oscillating function. Here I am not sure whether I have to use the condition $y(0)=\begin{pmatrix}
0\\ 
1
\end{pmatrix}$ ,which I did not use at all. 
Am I going in the right direction? If not can someone point  me in the right direction? Any better approach of tackling the problem will be appreciated. Thanks in advance for your time.
 A: Yes you are, here are some hints. An initial value problem of the type
$\frac{dy}{dt}=Ay,\quad\quad y(0)=y_0\in\mathbb{R}^n$
where $A$ is an $n\times n$ real diagonalisable matrix has a unique solution given by
$y(t)=\sum_{i=1}^n c_iv_ie^{\lambda_i t}\quad\quad\quad(*)$
where $\lambda_i$ denote the eigenvalues of $A$, $v_i$ denotes the eigenvector corresponding to $\lambda_i$ and $c_i\in\mathbb{C}$ denote constants picked such that $y(0)=y_0$. In addition, note the identities
$trA=\lambda_1+\lambda_2+\dots+\lambda_n$ and    $\quad detA=\lambda_1\lambda_2\dots\lambda_n$,
what do they imply about the eigenvalues of your matrix? Let me know if you would like me to elaborate any further. 

EDIT: $A$ is $2\times 2$, so at most it has $2$ distinct eigenvalues, $\lambda_1$ and $\lambda_2$. Since trace$(A)=0$ and trace$(A)=\lambda_1+\lambda_2$, we must have that $\lambda_2=-\lambda_1$.
Next, $\det(A)=\lambda_1\lambda_2=-\lambda_1^2>0$. The inequality is only satisfied if $\lambda_1$ is an imaginary number, that is $\lambda_1=\alpha i$ where $\alpha\neq0$ is some real number.
Since $A$ has two distinct eigenvalues, it is diagonalisable. Which implies that $y$ must be of the form of $(*)$. That is,
$$y(t)=c_1v_1e^{\lambda_1 t}+c_2v_2e^{\lambda_2 t}=c_1v_1e^{\alpha i t}+c_2v_2e^{-\alpha i t}=c_1v_1[\cos(\alpha t)+i \sin(\alpha t)]+c_2v_2[\cos(\alpha t)-i \sin(\alpha t)].$$
In other words, $y_1$ and $y_2$ are linear combinations of periodic functions, and thus must periodic themselves (unless they exactly cancel out leaving $y\equiv 0$, however you can rule this out since $y(0)=[0$ $1]^T$). 
