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 Exercise [21.02] 
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Joined: 26 Mar 2010, 04:39
Posts: 109
Post 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
Post Re: Exercise [21.02]
I think we can assume (via the Picard existence theorem, like you mention) that given an (ordinary) differential equation:


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

y(x_0)=C_0, \;\; y^{(1)}(x_0)=C_1, \;\; ... , \;\; y^{(n-1)}(x_0)=C_{n-1}

Singles out a unique solution to the equation (at least in some neighborhood of x_0). (Where C_0, C_1,,... etc. are constants).

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

y = x^5 - 20x^3 + 120x + A\cos{x} + B\sin{x}

we can, by adjusting A and B, set y(x_0) and y^\prime(x_0) to any values we choose, for any given x_0. 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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