Verify nth derivative satisfies differential equation I've been working my way through  an old series of maths books (An Analytical Calculus by Maxwell) and finally got stuck on a question midway through book 2 (of 4). If anyone could help that would be great (as there are quite a few like this following). The question is in two parts and I can get the first part. I figure that the first part is to be used in the solution of the second part (but could be wrong). We haven't covered ODEs yet, so any methods using theorems from ODEs shouldn't be used.
a) Prove that
$$
(x+1)\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+1}~\left\{(x+1)^n(x-1)^{n+1}\right\}
=(n+1)\left(\frac{\mathrm d}{\mathrm d x}\right)^n\left\{(x+1)^{n+1}(x-1)^n\right\} 
$$
--> I can do this part.
b) Prove also that the function
$$
\left(\frac{\mathrm d}{\mathrm d x}\right)^n~{(x+1)^{n+1}(x-1)^n}
$$
Satisfies the equation:
$$
(1-x^2)\left(\frac{\mathrm d}{\mathrm d x}\right)^2y-(1+x)\left(\frac{\mathrm d y}{\mathrm d x}\right)+(n+1)^2y=0
$$ --> I can't do this part.
Worst case I thought I could do it via a recurrence relation between terms in a power series in the DE, and then check the function satisfies it, but it gets messy fast. I can't find any similar solved problems anywhere!
The first part is solved via Leibniz:
$$
(x+1)\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+1}{(x+1)^n(x-1)^{n+1}}=\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+1}{(x+1)^{n+1}(x-1)^{n+1}}-(n+1)\left(\frac{\mathrm d}{\mathrm d x}\right)^n{(x+1)^n(x-1)^{n+1}}
=\left(\frac{\mathrm d}{\mathrm d x}\right)^n(n+1){(x+1)^{n+1}(x-1)^n}+\left(\frac{\mathrm d}{\mathrm d x}\right)^n(n+1){(x+1)^n(x-1)^{n+1}}-(n+1)\left(\frac{\mathrm d}{\mathrm d x}\right)^n{(x+1)^n(x-1)^{n+1}}=\left(\frac{\mathrm d}{\mathrm d x}\right)^n(n+1){(x+1)^{n+1}(x-1)^n}
$$
 A: It can be solved via multiple Leibniz and chain rules. It's a long proof and easy to make mistakes. Here are the main parts of it.
First, using similar techniques to part a):
$$
(x+1)\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+1}~{(x+1)^{n+1}(x-1)^n}=\left(\frac{\mathrm d}{\mathrm d x}\right)^{n}~{(x+1)^{n+1}(x-1)^n}[(n+1)x+n-1].
$$
Then using Leibniz:
$$
\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+2}~{(x+1)^{n+1}(x-1)^n(x-1)(x+1)}=(x^2-1)\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+2}~{(x+1)^{n+1}(x-1)^n}+(n+2)2x\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+1}~{(x+1)^{n+1}(x-1)^n}+(n+2)(n+1)\left(\frac{\mathrm d}{\mathrm d x}\right)^{n}~{(x+1)^{n+1}(x-1)^n}
$$
Then

*

*Expand the LHS using the chain rule and then the chain rule again on each of the two outputs

*Expand the second term on the RHS using similar techniques to part a), and

*Leave the third term alone.

This gives:
$$
(1-x^2)\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+2}~{(x+1)^{n+1}(x-1)^n}-(x+1)\left(\frac{\mathrm d}{\mathrm d x}\right)^{n+1}~{(x+1)^{n+1}(x-1)^n} = \left(\frac{\mathrm d}{\mathrm d x}\right)^{n}~{(x+1)^{n}(x-1)^{n-1}}[-(n+2)(n+1)(x-1)^2-2(n+1)(n+2)(x+1)(x-1)-n(n+1)(x+1)^2+2(n+2)(x+1)(x-1)+2(n+2)(n+1)x(x-1)+2(n+2)nx(x+1)-2(n+2)(n+1)(x+1)(x-1)+(n+2)(n+1)(x+1)(x-1)-(x+1)(x(n+1)+n-1)] 
$$
The above RHS rationalizes to:
$$
-(n+1)^2\left(\frac{\mathrm d}{\mathrm d x}\right)^{n}~{(x+1)^{n+1}(x-1)^n}
$$
Hence the result. A very long proof (at least by the standards of the other questions in the book) to solve something that looks like it might not be too difficult to prove!.
