I am trying to prove the identity;
$ \left[ n(n+1)-m(m+1) \right] j_n(x)j_m(x) = \frac{d}{dx}\left(x^2j_n'(x)j_m(x) - j_n(x)j_m'(x) \right). $
I began with the differential equations satisfied by $j_n(x)$ and $j_m(x)$, namely;
$ x^2j_n'' +2xj_n' + (x^2-n(n+1))j_n= 0, $
$ x^2j_m'' +2xj_m' + (x^2-m(m+1))j_m= 0,$
where the argument of $x$ in each function is implied.
These can be rearranged to give;
$ n(n+1)j_n = x^2 j_n'' +2xj_n' +x^2 j_n , $
$ m(m+1)j_m = x^2 j_m'' +2xj_n' +x^2 j_m . $
Returning to our initial expression;
$ [n(n+1) - m(m+1)]j_n(x)j_m(x) $
$ = j_m(x)[n(n+1)j_n(x)] - j_n(x)[m(m+1)j_m(x)] $
$ = j_m(x)[x^2j_n''(x) + 2xj_n'(x) +x^2j_n(x)] - j_n(x)[x^2j_m''(x) +2xj_m'(x) +x^2 j_m(x) ] $
$ = x^2j_n''(x)j_m(x) -x^2j_n(x)j_m''(x)+2xj_n'(x)j_m(x)-2xj_n(x)j_m'(x) +x^2j_n(x)j_m(x) -x^2j_n(x)j_m(x) $
$ = x^2[j_n''(x)j_m(x)-j_n(x)j_m''(x)] +2x[j_n'(x)j_m(x) - j_n(x)j_m'(x)] $
$ u = x^2 \implies u' =2x $
$ v = j_n'(x)j_m(x) - j_n(x)j_m'(x) \implies v' = j_n''(x)j_m(x) - j_n(x)j_m''(x) $
Hence our expression is;
$uv' + u'v $ $ = (uv)' $
by the product rule.
$ = \frac{d}{dx}[(x^2)(j_n'(x)j_m(x) - j_n(x)j_m'(x))] $
$ = \frac{d}{dx} \Bigl(x^2 j_n'(x)j_m(x) - x^2 j_n(x)j_m'(x) \Bigr) $
Where did it go wrong?