Reciprocal solutions of a differential equation

Question

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

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

Answer 1

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.

Answer 2

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 $c$, $\text{arsinh}$).