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Differentiable function in $\mathbb{R}$ for which $f''(x) + e^x f(x)=0$ for every $x$. Prove that $f(x)$ is bounded as $x \rightarrow +\infty$

I have tried a few stuff but they didnt work out, for example i noticed that the function has infinite max and min as $x \rightarrow +\infty$ but thats still not enough to prove it, any ideas?

baudolino
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Plom
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1 Answers1

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The general solution of your differential equation is $$ f(x) = A J_0(2 e^{x/2}) + B Y_0(2 e^{x/2})$$ where $J_0$ and $Y_0$ are Bessel functions of the first and second kinds. Their asymptotics are known.

EDIT: if $R(x) = f'(x)^2 + e^x f(x)^2$, show that $R'(x) = e^x f(x)^2 \le R(x)$, and thus $\dfrac{d}{dx} \log R(x) \le 0$.

Robert Israel
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    i appreciate the answer but i cant understand anything since i dont know what a bessel function is... is there another way of showing this? i think there must be cuz my book doesnt get into DE solving at all, just derivatives and basic theorems about them... thx again – Plom Aug 18 '13 at 05:29