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Math Help - Higher order derivatives

  1. #1
    kin
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    Higher order derivatives

    please show me how to solve the following.
    i have tried differentiation twice , but i cant get the same results..
    Last edited by mr fantastic; September 23rd 2009 at 10:29 PM.
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  2. #2
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    Quote Originally Posted by kin View Post
    please show me how to solve the following.
    i have tried differentiation twice , but i cant get the same results..
    First, I believe there's a typo. I think it should be y = \frac{\sin x}{1-x^2} and not y = \frac{\sin x}{(1-x)^2}.

    If you consider

    (1-x^2) y = \sin x

    then differentiating both sides wrt x twice and adding the same expression gives

     <br />
\left( (1-x^2)y\right)'' + (1-x^2)y = 0.<br />
Expanding gives (1-x^2)y'' - 4x y' - (1+x^2)y = 0 - the first result.

    If we let u = (1-x^2)y then we have the hierarchy of differential equations

     <br />
u^{(n+2)} + u^{(n)} = 0\;\;(1)<br />
.

    By induction we can show that

     <br />
u^{(n)} = (1-x^2) y^{(n)} - 2n x y^{(n-1)} - n(n-1)y^{(n-2)}<br />

    and at x = 0

     <br />
u^{(n)} = y^{(n)} - n(n-1)y^{(n-2)}.<br />

    Substitution into (1) leads to the second result

     <br />
y^{(n+2)} -(n^2+3n+1) y^{(n)} - n(n-1)y^{(n-2)} = 0<br />
.
    Last edited by Jester; September 23rd 2009 at 03:20 PM.
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  3. #3
    kin
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    Quote Originally Posted by Danny View Post
    First, I believe there's a typo. I think it should be y = \frac{\sin x}{1-x^2} and not y = \frac{\sin x}{(1-x)^2}.
    this is reallly important , thanks danny....
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