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Let $q$ be a real valued non-trivial solution solution of $$ y'' +A(x)y = 0 \text{ on } a<x<b, $$ and let $w$ be a real valued non-trivial solution of $$ y'' + B(x)y = 0 \text{ on } a<x<b. $$ Here $A$ and $B$ are real valued continuous functions satisfying $$ B(x)>A(x) \text{ for } a<x<b. $$ How to show that if $x_1$ and $x_2$ are successive zeros of $q$ on $(a,b)$, then $w$ must vanish at some point $p \in (x_1, x_2)$?

Partial answer: Let $q, w>0$ on $(x_1, x_2)$,then with $(wq'-qw')'= (B-A)qw$, and by integration from $x_1$ to $x_2$ we get $w(x_2)q'(x_2)-w(x_1)q'(x_1)> 0$. Somehow I want to show that that $q'(x_1)< 0$ or $q'(x_2)>0$, which will then contradict $q > 0$ on $(x_1, x_2)$

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You have done all the hard work. Since $q>0$ on $(x_1,x_2)$ and $q(x_1)=q(x_2)=0$, uniqueness implies that $q'(x_1)>0$ and $q'(x_2)<0$. Finally, since $w>0$ on $(x_1,x_2)$, $w(x_i)\ge0$, $i=1,2$. Then $$ w(x_2)q'(x_2)-w(x_1)q'(x_1)\le0, $$ which gives you a contradiction.

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How do you deduce that $q'(x_1)>0, q'(x_2)<0$? – Mercy King Sep 6 '12 at 13:36
Since $q(x)>0=q(x_1)$ for $x>x_1$, $(q(x)-q(x_1))/(x-x_1)>0$ for $x>x_1$. Passing to the limit as $x\to x_1$ gives $q'(x_1)\ge0$. If $q'(x_1)=0$, by uniqueness of solution, $q\equiv0$. Since $q$ is nontrivial, it must be $q'(x_1)>0$. A similar argument gives $q'(x_2)>0$. – Julián Aguirre Sep 6 '12 at 13:56

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