Differential equation help

For the discussion of math. Duh.

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Differential equation help

Hi all,

I've got a second order D.E., and the way that I would normally solve it gives an answer different to what I think I should get.
$\ddot{\phi} = c\phi$
Where c is some combined constant. So, you just solve the auxiliary equation,
$k^2 - c = 0$
$k_1 = \sqrt{c},\;\; k_2 = - \sqrt{c}$
So,
$\phi = Ae^{\sqrt{c}x} + Be^{-\sqrt{c}x}$

Is the answer I'd get. But the answer I'm supposed to be getting only has the [imath]Be^{-\sqrt{c}x}[/imath] term, and I can't see how you would lose the other bit. Am I doing it totally wrong?

Any help would be appreciated.

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Re: Differential equation help

Do you have any initial conditions?