# 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) $\{k:k<-1\}.$

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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

$$|e^{(a+ib)x}|= | e^{ax}|| e^{ibx}|=e^{ax} .$$

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Thanks a lot sir for the detailed explanation. I have got it. – learner Dec 14 '12 at 15:07
@learner: You are welcome. Glad to assist. – Mhenni Benghorbal Dec 14 '12 at 15:17

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.

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