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Let $y(x)$ be the solution of the differential equation $\frac{d^2y}{dx^2} - y = 0$ such that $y(0) = 2$ and $y'(0) = 2\alpha$. Find all values of $\alpha\in[0,1)$ such that the infimum of the set { $y(x) | x\in \mathbb{R} $} is greater than or equal to $1$.

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up vote 4 down vote accepted

The solution to the differential equation is

$$ y(x)=c_1 \cosh x + c_2 \sinh x $$

and imposing the initial conditions we have

$$ y(x)=2 (\cosh x + \alpha \sinh x)=2\sqrt{1-\alpha^2}\cosh(x+A) $$

for an opportune value of $A$ ($=\tanh^{-1}\alpha$).

Now should be obvious how to go on and determine the required values for $\alpha$.

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We can sole this equation, whose solution is $f_{\alpha}(x)=(1+\alpha)e^x+(1-\alpha)e^{—x}$. We have $\lim_{|x|\to +\infty}f_{\alpha}(x)=+\infty$ if $\alpha\in [0,1)$, hence the minimum is attained, and necessarily at a critical point. At such a point, we should have $(1+\alpha)e^x-(1-\alpha)e^{—x}=0$ hence $e^{2x}=\frac{1-\alpha}{1+\alpha}$. We have $$f_{\alpha}\left(\frac 12\log\frac{1-\alpha}{1+\alpha}\right)=\sqrt{\frac{1-\alpha}{1+\alpha}}(1+\alpha)+(1-\alpha)\sqrt{\frac{1+\alpha}{1-\alpha}}=2\sqrt{1-\alpha^2}.$$ Hence $\inf_{x\in\Bbb R}f_{\alpha}(x)\geq 1$ is equivalent to $2\sqrt{1-\alpha^2}\geq 1$, $1-\alpha^2\geq\frac 14$ so $\alpha^2\leq \frac 34$ and finally $\alpha\leq \frac{\sqrt 3}2$.

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