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 $\displaystyle y = \frac{\sin x}{1-x^2}$ 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
$.