Reciprocal solutions of a differential equation 
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
$$u''-\frac{2(u')^2}{u}+(\frac{b'(x)}{b(x)})(u')+(\frac{a^2}{[b(x)]^2})(u(x))=0\,.$$
 A: You got so close. Suppose that $y_1=u$ is a solution, then $y_2=1/u$ must also be a solution. Substituting this, we obtain
$$ u'' -2\frac{(u')^2}{u} + \frac{b'}{b}u' + \frac{a^2}{b^2}u = 0 $$
But we already know that 
$$ u'' + \frac{b'}{b}u' = \frac{a^2}{b^2}u $$
From the given assumption. Therefore
$$ 2\frac{a^2}{b^2}u - 2\frac{(u')^2}{u} = 0 \implies \frac{u'}{u} = \pm\frac{a}{b} $$
Integrating both sides gives
$$ u(x) = c\exp\left(\pm \int \frac{a}{b(x)} dx\right) $$
where $c$ is some arbitrary constant. Taking the positive and negative signs to be distinct solutions, we can put together the general solution
$$ y(x) = c_1 \exp\left(\int \frac{a}{b(x)}dx \right) + c_2 \exp\left(-\int \frac{a}{b(x)}dx\right) $$
which is indeed a linear combination of two reciprocal functions.
Hope this helps.
A: Hint:
This is probably not the expected answer, but in fact you can integrate the equation.
$$b(x)y''(x)+b'(x)y'(x)=(b(x)y'(x))'=\frac{a^2y(x)}{b(x)},$$ and multiplying by $b(x)y'(x)$,
$$b(x)y'(x)(b(x)y'(x))'=a^2y(x)y'(x).$$
By integration,
$$(b(x)y'(x))^2=a^2y^2(x)+c.$$
This is a separable equation,
$$\frac{y'(x)}{\sqrt{a^2y^2(x)+c}}=\pm\frac1{b(x)}.$$
$$\frac1a\text{arcosh}\frac{ay(x)}{\sqrt c}=\pm\int\frac{dx}{b(x)}$$ (for negative $c$, $\text{arsinh}$).
