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For a smooth function $f: (-\pi/2,\pi/2) \to \mathbb{R} $, if \displaystyle\frac{|f''(x)|}{\sqrt{(1+f'(x))^3}} = \cos{x}, and f(0) = f'(0) = 0, f''(0) = 1, f''(-\pi/2) = 0, and f''(\pi/2) = 0. What is the function $f(x)$?

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    @Henning, the boundary conditions are met no matter what: $f''$ must vanish at $\pm\pi/2$ because $\cos x$ is zero there.2011-09-18

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This is a second order diferential equation with three initial conditions (let's forget by now about the boundary conditions), while usually two are enough to have existence and uniqueness. However, from the equation and $f(0)=f'(0)=0$, it follows that $|f''(0)|=1$, which is compatible with the third condition $f''(0)=1$. Since $f''(0)>0$, the second derivative of the solution will be positive in a neighbourhood of $0$ ; thus, in a first step, we can forget abot the absolute value.

Let $f´=y$. The problem becomes $ (1+y)^{-3/2}y'=\cos x,\quad y(0)=0. $ This is a first order equation in separate variables, whose solution is $ y=\frac{4}{(2-\sin x)^2}-1. $ Then $ f(x)=-x+\int_0^x\frac{4\,dt}{(2-\sin t)^2}, $ and $ f''(x)=y'(x)=\frac{8\cos x}{(2-\sin x)^3}. $ Since $f''>0$ on $(-\pi/2,\pi/2)$ and $f''(-\pi/2)=f''(\pi/2)=0$, $f$ is the solution you are looking for. Below is the graph of $f$, $f'$ and $f''$.

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