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Exercise [21.02]
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Joined: 26 Mar 2010, 04:39
Posts: 109
Exercise [21.02]
Here is exercise 21.2

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21 Jul 2010, 04:43

Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [21.02]
I think we can assume (via the Picard existence theorem, like you mention) that given an (ordinary) differential equation:

Where is analytic (actually this is a stronger condition than required, but it's all we need here), then specifying a set of initial conditions:

Singles out a unique solution to the equation (at least in some neighborhood of ). (Where ,... etc. are constants).

Given that, it's a simple matter to show that for the general solution

we can, by adjusting A and B, set and to any values we choose, for any given . Thus, the above general solution includes every possible solution, and so is the most general possible.

12 Aug 2012, 09:20
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