# Higher order derivatives

• Sep 23rd 2009, 08:38 AM
kin
Higher order derivatives
please show me how to solve the following.
i have tried differentiation twice , but i cant get the same results..
http://f20.yahoofs.com/hkblog/_9eQUJ..._____DH_9lrq2O
• Sep 23rd 2009, 03:06 PM
Jester
Quote:

Originally Posted by kin
please show me how to solve the following.
i have tried differentiation twice , but i cant get the same results..
http://f20.yahoofs.com/hkblog/_9eQUJ..._____DH_9lrq2O

First, I believe there's a typo. I think it should be $\displaystyle y = \frac{\sin x}{1-x^2}$ and not $\displaystyle y = \frac{\sin x}{(1-x)^2}$.

If you consider

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

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

$\displaystyle \left( (1-x^2)y\right)'' + (1-x^2)y = 0.$ Expanding gives $\displaystyle (1-x^2)y'' - 4x y' - (1+x^2)y = 0$ - the first result.

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

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

By induction we can show that

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

and at $\displaystyle x = 0$

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

Substitution into (1) leads to the second result

$\displaystyle y^{(n+2)} -(n^2+3n+1) y^{(n)} - n(n-1)y^{(n-2)} = 0$.
• Sep 23rd 2009, 07:48 PM
kin
Quote:

Originally Posted by Danny
First, I believe there's a typo. I think it should be $\displaystyle y = \frac{\sin x}{1-x^2}$ and not $\displaystyle y = \frac{\sin x}{(1-x)^2}$.

this is reallly important , thanks danny....(Wink)