> 
> I need to show that if $a$ is a constant and $b(x)$ is a function, then
> $$y''+\frac{b'(x)}{b(x)}y'-\frac{a^2}{[b(x)]^2}y=0$$
> has a pair of linearly independent solutions which are reciprocal and then find them.
> 
> 
> 

I would have thought I could just substitute in $y=u(x)+\dfrac{1}{u(x)}$ but I seem to get nowhere.

Anyway, we have $y'=u'-\dfrac{u'}{u^2}$ and $y''=u''-\dfrac{u''}{u^2}+\dfrac{2(u')^2}{u^3}$. Therefore, we get $$u''-\frac{u''}{u^2}+\frac{2(u')^2}{u^3}+(\frac{b'(x)}{b(x)})(u'-\frac{u'}{u^2})-(\frac{a^2}{[b(x)]^2})(u(x)+\frac{1}{u(x)})=0\,.$$

tidying gives

$$-\frac{u''}{u^2}+\frac{2(u')^2}{u^3}+(\frac{b'(x)}{b(x)})(-\frac{u'}{u^2})-(\frac{a^2}{[b(x)]^2})(\frac{1}{u(x)})=0\,.$$

multiply by $-u^2$ to get

$$u''-\frac{2(u')^2}{u}+(\frac{b'(x)}{b(x)})(u')+(\frac{a^2}{[b(x)]^2})(u(x))=0\,.$$