# Isn't this a problem when evaluating the auxiliary equation?

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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$$e^{\alpha x}\xrightarrow [x\to -\infty]{}0\Longleftrightarrow \alpha >0$$ – DonAntonio Sep 24 '12 at 12:43
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. – Did Sep 24 '12 at 12:50
What exactly did you enter in Wolfram Alpha? Because I get "$0$ assuming $\alpha>0$" – celtschk Sep 24 '12 at 12:52
@KorganRivera: That last comment deserves a downvote of its own -- here you go. – Henning Makholm Sep 24 '12 at 13:50
@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. – celtschk Sep 24 '12 at 13:53

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.