Higher order derivatives

**kin** · September 23rd 2009, 08:38 AM

please show me how to solve the following.
i have tried differentiation twice , but i cant get the same results..

**Danny** · September 23rd 2009, 03:06 PM

Originally Posted by kin

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

$ \left( (1-x^2)y\right)'' + (1-x^2)y = 0. $ 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

$ u^{(n+2)} + u^{(n)} = 0\;\;(1) $ .

By induction we can show that

$ u^{(n)} = (1-x^2) y^{(n)} - 2n x y^{(n-1)} - n(n-1)y^{(n-2)} $

and at $x = 0$

$ u^{(n)} = y^{(n)} - n(n-1)y^{(n-2)}. $

Substitution into (1) leads to the second result

$ y^{(n+2)} -(n^2+3n+1) y^{(n)} - n(n-1)y^{(n-2)} = 0 $ .

**kin** · September 23rd 2009, 07:48 PM

Originally Posted by Danny

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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