Originally Posted by

**MonroeYoder** actually you're doing just fine =). After differentiating once we have n=1. So the $\displaystyle (1-x^2)v^{(3)}$ corresponds to the $\displaystyle (1-x^2)v^{n+2}$

Then $\displaystyle -2xv''$ corresponds to $\displaystyle -2xv^{n+1}$

Now the part you prolly didnt realize, since $\displaystyle n=1$, we have $\displaystyle 2(n-1)(xv''+v')=2(1-1)(xv''+v')=0$

And now $\displaystyle 2nv' = 2v'$ which corresponds to $\displaystyle n(n+1)v^{(n)} = 1(1+1)v^{(1)}=2v'$

So all of your terms follow the pattern perfectly... now just keep differentiating