Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ...

share|improve this question
    
What is $u$ here? –  Emmad Kareem 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
1  
...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

1 Answer 1

up vote 1 down vote accepted

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: $$ 2 p' u' - \frac{2 p u'^2}{u} = 0. $$

share|improve this answer
    
Ugh. This is exactly what I did, but I made a calculation error which made me think it wouldn't work. –  ro44 Oct 30 '12 at 23:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.