Consider the second order linear differential operators $L$ and $\bar{L}$ given by $$ L = p(x) \frac{d^2}{dx^2} + q(x) \frac{d}{dx} + r(x), \ \ \bar{L} = \frac{d^2}{dx^2}p(x) - \frac{d}{dx} q(x) + r(x) $$ Show that, for all pairs of functions $f(x), g(x)$ we have $$ fL[g]-g\bar{L}[f] = \frac{d}{dx}( \mathcal{B}[f,g]) $$ where $\mathcal{B}[f,g]$ is some quantity that depends on $f, f', g$ and $g'$
So I've attempted this by just working through the expression, but I'm unable to get something that just depends on $f, f', g$ and $g'$? I'm not sure if I'm differentiating wrong or just misunderstanding the question. I have found similar things trying to understand this problem but I cannot seem to find this exact identity anywhere, only with symmetric operators, which is not the case here. But, here's what I have:
\begin{align} \bar{L} &= \frac{d^2}{dx^2}p(x) - \frac{d}{dx} q(x) + r(x) \\ \\ &= p(x) \frac{d^2}{dx^2} + (2p'(x) - q(x))\frac{d}{dx} + (q''(x) - q'(x) + r(x)) \end{align} So, we get \begin{align} fL[g]-g\bar{L}[f] &= fp(x)\frac{d^2 g}{dx^2} + fq(x)\frac{dg}{dx} + fr(x)g \\ &- gp(x)\frac{d^2 f}{dx^2} - 2gp'(x)\frac{df}{dx} + gq(x)\frac{df}{dx} - gp''(x)f + gq'(x)f - gr(x)f \\ \\ &= pfg'' + qfg' + rfg - pf''g - 2p'f'g + qf'g - p''fg + q'fg - rfg \\ \\ &= pfg'' + qfg' - pf''g - 2p'f'g + qf'g - p''fg + q'fg \end{align}
and, I'm pretty much stuck on where to go from here...
I've just been trying to find different ways to simplify and the furthest I've got is $$ fL[g]-g\bar{L}[f] = p \frac{d}{dx}[f'g-fg'] + q \frac{d}{dx}[fg] + g(q'f - p''f - 2p'f') $$
But, this doesn't appear necessarily useful to me so I'm unsure quite what to do here.
The calculation becomes more manageable if we write $$ L=L_p+L_q+L_r, \tag{1} $$ where $$ L_p:=p(x)\frac{d}{dx^2},\quad L_q:=q(x)\frac{d}{dx},\quad L_r:=r(x), \tag{2} $$ and similarly for $\bar{L}$. Thus, in decreasing order of simplicity, we have $$ fL_r[g]-g\bar{L}_r[f]=frg-grf=0, \tag{3} $$ $$ fL_q[g]-g\bar{L}_q[f]=fqg'+g(qf)'=(qfg)', \tag{4} $$ and \begin{align} fL_p[g]-g\bar{L}_p[f]&=fpg''-g(pf)''=fpg''+(fp)'g'-g'(pf)'-g(pf)'' \\ &=[(fp)g']'-[g(pf)']'=[pfg'-(pf)'g]'. \tag{5} \end{align} Collecting all pieces together, we finally obtain $$ fL[g]-g\bar{L}[f]=\frac{d}{dx}[pfg'-(pf)'g+qfg]. \tag{6} $$