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I was thinking about the following problem that says:

Let $a,b\in \mathbb{R}$.Let $y=(y_1,y_2)^{t}$ be a solution of the system of $y'_1=y_2$,$y'_2=ay_1+by_2$. Then which of the following options is correct?

Every solution $y(x)\rightarrow 0$ as $x\rightarrow \infty$ if

(a) $a<0,b<0$

(b) $a<0,b>0$

(c) $a>0,b>0$

(d) $a>0,b<0.$

My attempts: From the second equation we see, $y_1''=ay{_1}'+by{_2}'=ay_2+by{_2}'$.The auxiliary equation of this D.E. is $m^2-bm-a=0$ and hence $m_1=(b+\sqrt(b^2+4a))/2$,$m_2=\frac{b-\sqrt{b^2+4a}}{2}$. Hence the solution is given by $y=c_1e^{m_1x}+c_2e^{m_2x}$.Now,$y_2$ will tend to 0 for any choice of $c_1$ and $c_2$ if and only if the real parts of m1 and m2 are negative(Thanks to Antonio Vargas sir).Now $m_2<0$ gives $b<\sqrt(b^2+4a)$ and so $b^2<b^2+4a$ and hence $a>0$. From here, i could not progess. Please help. Thanks in advance for your time.

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What you found for $y_2$ will tend to $0$ for any choice of $c_1$ and $c_2$ if and only if the real parts of $m_1$ and $m_2$ are negative. – Antonio Vargas Dec 9 '12 at 8:53
up vote 1 down vote accepted


If $b>0$, the solution will either be of the form $e^\frac{b+\Delta}{2}$ or $e^\frac{b+i\Delta}{2}$ depending on whether the discriminant $b^2+4a$ is negative or positive. Both will tend to $\infty$ as $x \to \infty$. So, $b>0$ is not possible.


In this case, let $b = -\beta, \beta>0$. If $a>0$, the discriminant is always positive. $m_1 = \frac{\sqrt{\beta^2+4a}-\beta}{2} >0$. So, $e^{m_1}\to \infty , x\to \infty$. If $a<0$, it is possible that the discriminant is negative. Then, the solution is of the form $e^{-\beta}(A\cos{\alpha t} + B \sin{\alpha t}), \alpha,\beta>0$. This will tend to zero as $x \to \infty$. So, the correct answer is $a<0,b<0$

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@dexter04.Thank you sir for the explanation. I have got it. – learner Dec 9 '12 at 9:59
@learner: my pleasure – dexter04 Dec 9 '12 at 10:00

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