Problem related to a differential equation I am stuck with the following problem:
Let $Y(x)=(y_{1}(x),y_{2}(x))$ and let $A$ is given by $$\begin{pmatrix}
-3 &1 \\ 
 k& -1
\end{pmatrix}.$$
Further, let $S$ be the set of values of $k$ for which all the solutions of the system of equations $Y'(x)=AY(x)$ tend to $0$ as $x$ tends to $\infty.$ Then $S$ is given by: 
(a) $\{k:k\leq -1\}$
(b) $\{k:k\leq 3\}$
(c) $\{k:k<3\}$
(d) $\{k:k<-1\}.$
Please help. Thanks in advance for your time.
 A: Recalling the general solution of your equation,
$$ Y(x)=c_1 v_1 e^{\lambda_1 x} + c_2 v_1 e^{\lambda_2 x}, $$
where $\lambda_1,\lambda_2$ are the eigenvalues and $v_1,v_2$ are the eigenvectors. Now, what matters in your problem is the eigenvalues, since you have the condition that the solution $Y(x)$ goes to $0$ as $x\to \infty$. This requires the following condition on the eigenvalues 
$$ Real(\lambda_1) < 0, Real( \lambda_2 ) < 0 . $$
The eigenvalues of your matrix are
$$ \lambda_1=-2+\sqrt{k+1}, \,, \lambda_2=-2-\sqrt{k+1}, $$
and 
$$ \lambda_1=-2+\sqrt{k+1}<0, \,, \lambda_2=-2-\sqrt{k+1}<0.  $$
Work this out and you will find the answer is $(c)$? 
Note: When you take the limit of $e^{(a+ib)x}$ as $x \to \infty$, what matters is the real part, since
$$ |e^{(a+ib)x}|= | e^{ax}|| e^{ibx}|=e^{ax} . $$
A: Do you know how to express the solutions in terms of the eigenvalues and eigenvectors of $A$? If you know that formula, you can see what condition you need on the eigenvalues for the solutions to tend to zero, and then you can work out the values of $k$ for which the eigenvalues satisfy that condition. 
