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Prove that if $f$ is differentiable on $[a,b]$ and if $f(a)=f(b)=0$ then for any real $\beta$ there is an $x \in (a,b)$ such that $\beta \cdot f(x)+f'(x)=0$. (Using rolle's theorem)

My attempt:

Using Rolle's theorem we can say that there exists some $c \in (a,b)$, where $f'(c)=0$ . Therefore one factor in the expression $\beta \cdot f(x)+f'(x)=0$ is $0$ at $c$ but I am unable to prove that the other factor will simultaneously be $0$ at $c$.

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Hint: Apply Rolle's Theorem to $g(x) = e^{\beta x} f(x)$.

$g(a) = g(b) = 0$, therefore there is a $c \in (a, b)$ such that $ g'(c) = 0$. The conclusion follows with $g'(x) = e^{\beta x}(\beta f(x) + f'(x))$ since $e^{\beta x}$ is never zero.

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  • $\begingroup$ Thanks I got the proof now $\endgroup$ – user383555 Nov 19 '16 at 22:42
  • $\begingroup$ This is pretty slick and obvious in retrospect. And I suppose this is a common enough trick in differential equations to occur to someone naturally. However, is there a particular reason why a student who just encountered Rolle's Theorem should think of this trick? $\endgroup$ – Robert Wolfe Dec 21 '18 at 5:48

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