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The auxiliary equation of a differential equation is of the form

$eK^{\alpha x}(am^2+bm+c)=0$

$Ke^{\alpha x}$ can't be zero so long as K isn't zero. But

$lim_{x \to -\infty}Ke^{\alpha x}=0$

Isn't that a problem when evaluating the auxiliary equation?

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    $$e^{\alpha x}\xrightarrow [x\to -\infty]{}0\Longleftrightarrow \alpha >0$$2012-09-24
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    *What does that mean?* That W|A is not supposed to bring mathematical understanding, only to perform computations (and even, of a certain kind). In the case at hand, the quadratic might be $ax^2+bx+c$, not what you wrote.2012-09-24
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    What exactly did you enter in Wolfram Alpha? Because I get ["$0$ assuming $\alpha>0$"](http://www.wolframalpha.com/input/?i=lim_x-%3E-infinity+e^%28alpha+x%29&dataset=&asynchronous=false&equal=Submit)2012-09-24
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    @did I have no idea what you just said. O_O2012-09-24
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    @celtschk I...huh. That's weird. It gave me (0) last time. Anyway, isn't this a problem when evaluating the auxilary equation in a 2nd-order differential equation?2012-09-24
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    @KorganRivera: That last comment deserves a downvote of its own -- here you go.2012-09-24
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    @KorganRivera What part of my last comment is difficult to understand?2012-09-24
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    @KorganRivera: Since I don't know the 2nd order differential equation you are trying to solve, or how this equation relates to it, I cannot say for sure, but I'd bet no. For any finite $x$ and finite $\alpha$, either the polynomial or $K$ must be $0$ in order to solve that equation. So unless a legitimate solution of the differential equation would correspond to infinite $x$ or infinite $\alpha$, there should not be a problem.2012-09-24
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    Note: my comment above addressed the first version of this question, which has since passed through some rather odd formulations.2012-10-06

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No, it is not a problem, since the function $x \mapsto Ke^{\alpha x}$ is never zero for $x\in\mathbb R$, so the other one has to be a constant zero function.