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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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    Note: my comment above addressed the first version of this question, which has since passed through some rather odd formulations.2012-10-06

1 Answers 1

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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.