# $u,1/u$ are solutions to $(p(x)y')'+q(x)y=0$, show $u$ solves a first-order diff equation

I'm struggling with this question: we have two differentiable functions $p, q$ in some interval $I$ and the equation $(p(x)y')'+q(x)y=0$, with given solutions $u$ and $1/u$. We're asked to show $u$ solves a first-order differential equation.

What I tried: I tried two things: plugging the solutions to the equation and differentiating, and plugging the solutions and integrating (to get rid of the 2nd order derivative). It didn't get me very far, though ...

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What is $u$ here? – NoChance Oct 30 '12 at 23:35
Abel's Identity should help you out. – user12477 Oct 30 '12 at 23:37
@EmmadKareem: It's a function $u(x)$ that solves the equation – ro44 Oct 30 '12 at 23:39
@user12477: Thanks, but I think the solution is meant to be more elementary (we haven't learned much at all about 2nd order equations yet)! – ro44 Oct 30 '12 at 23:42
...I think you were on the right track in the first place. Since $u$ and $u^{-1}$ are both solutions, we have two separate equations involving $u''$. Eliminating $u''$ from these leaves an equation involving just $u$ and $u'$ (and coefficients depending on $p,p',q$) - in other words exactly what you've been asked to find! You don't need to integrate. – user12477 Oct 30 '12 at 23:49

We know $$\left(p u'\right)' + qu = p u'' + p' u' + qu = 0$$ and $$\left[p \left(\frac{1}{u}\right)'\right]' + \frac{q}{u} = \frac{2 p u'^2}{u^3} + \frac{p u''}{u^2} - \frac{p' u'}{u^2}+\frac{q}{u} = 0.$$ Multiply the second equation by $u^2$: $$\frac{2 p u'^2}{u} + p u'' - p' u' + q u = 0.$$ Subtract this equation from the first equation: $$2 p' u' - \frac{2 p u'^2}{u} = 0.$$