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the fucnction y(x) implement the Equation y''-xy=0.

In addition, know that y(0)=0 and y'(o)=1.

find the value of y(0)^(n), which means the value of 0 in the nth derivative.

Thanks.

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-1 Please try to post readable questions. –  Rasmus Feb 28 '11 at 14:45
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Don't post your questions in the imperative mode; if you have a question, please ask, don't give orders. Also: while your native language is likely not English, and so the English part may be excused, you could at least make some effort in getting the mathematical notation correct and legible. –  Arturo Magidin Feb 28 '11 at 16:14
    
(downvote removed after Arturo's edit) –  Rasmus Feb 28 '11 at 16:21
    
what do you mean by asking and not giving orders? you know, after a while I will eventually learn how to write all the terms correctly, but you should learn from Ross's comments in my last posts about constructive criticism. –  user6163 Feb 28 '11 at 16:57
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@Nir. For politeness, you should word a post like "How does one show that this happens..." instead of "Show that this happens." The first way is viewed as asking a question, where the second way is viewed as an imperative or giving someone a required assignment. –  JavaMan Feb 28 '11 at 17:10
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1 Answer

up vote 4 down vote accepted

Add xy to both sides (To get $y'' = xy$), and start taking derivatives. A pattern emerges.

Edit: You should get:

$y^{(3)} = y + xy'$

$y^{(4)} = 2y' + xy''$

$y^{(5)} = 3y'' + xy'''$

$y^{(6)} = 4y^{(3)} + xy^{(4)}$

$y^{(7)} = 5y^{(4)} + xy^{(5)}$

and so on...

Then, ask yourself what happens when you start evaluating each of these terms at $0$.

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Notice that I edited my equation. I did the calculation, and then wrote it without looking at my work! It should be correct now. –  JavaMan Feb 28 '11 at 14:53
    
Thank you so much, I'll be working on it. –  user6163 Feb 28 '11 at 15:03
    
ok, so no success with this. what should I do with the other data? –  user6163 Feb 28 '11 at 15:48
    
@Nir: If you want $y^{(n)}(0)$ you can just plug in your conditions to get $y^{(3)}$ and so on –  Ross Millikan Feb 28 '11 at 15:53
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