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Particular cases show that this is the case. E.g. free fall $\dfrac{d^2 x}{d t^2} = -g$, and the simple harmonic oscillator $\dfrac{d^2 x}{d t^2} + \omega^2x = 0$.

I can see why this is so physically, as in the above cases, but not in the general sense.

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    Am I missing something here? Surely in solving $\frac{\mathrm d^2x}{\mathrm dt^2}=-g$ you would need 2 initial conditions, equal to the order. Intuitively, if you want to know the position of a free falling object at a certain time what do you need to know except for the acceleration which is given by the equation? The position at a given $(x,t)$ and the velocity at a given $(x,t)$, right? As far as I know this isn't any different for the simple harmonic oscillator or other linear ODEs.2012-12-31
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    @Ron, I want to make sure you know about the "accept an answer" feature of this site. There is a checkmark to the left of each answer to your question, which you can use to mark the answer that was most helpful to you. It may be too early to do so for the present question, but you should accept some answers to your older questions. For details, see [Why should we accept answers?](http://meta.math.stackexchange.com/questions/3399/why-should-we-accept-answers_)2013-01-01

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