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Assume that $p(x)<0$ for all real numbers and y(x) is a solution of the DE $$y'+p(x)y=0$$ that is not identically zero. I need to prove that y can cross the x axis at most once.

I don't understand the answer below, any other help, please?

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up vote 3 down vote accepted

Rough idea: between two consecutive zeroes of $y$, $y'$ must vanish. But $y'=-p(x)y(x)$, which has constant sign between consecutives zeroes of $y$.

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@ siminore how do I know that Y has a zero or 2 zeros as you claim? –  Klara Dec 12 '12 at 12:47
    
Oh, you do not know. But you asked to prove the existence of at most one zero. I gave you a hint to prove that two zeroes are impossible, hence you can have either no zero or one zero. –  Siminore Dec 12 '12 at 13:25
    
@ siminore thank you for responding. You are saying that between consecutive zeros of y, y'=0 in all of the interval, or just at the zeros? Please be patient with me. –  Klara Dec 12 '12 at 13:39
    
@ siminore, correct me if I'm wrong, but I was thinking that in between the two zeros there is an extremum and there y'=0 but our DE says that y' cannot be zero in between zeros. Is this right? –  Klara Dec 12 '12 at 13:48
    
By Rolle's theorem, $f(a)=f(b)=0$ implies the existence of a critical point between $a$ and $b$. Then you can derive a contradiction with the DE. –  Siminore Dec 12 '12 at 17:34
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