The continuous function and twice differentiable function $f: \mathbb{R}\rightarrow\mathbb{R}$ satisfies $2007^{2}\cdot f(x)+f''(x)=0$. Prove that there exist two such real numbers $k$ and $l$ such that $f(x)=l\cdot\sin(2007x)+k\cdot\cos(2007x)$.
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Tags: function, calculus, derivative, real analysis, real analysis unsolved
13.03.2007 13:15
Just a second order differential equation...
13.03.2007 14:00
refer this http://www.mathlinks.ro/Forum/viewtopic.php?t=136624
14.03.2007 01:34
That's a terrible contest problem. It's trivial to anyone with appropriate advanced material (uniqueness theorems for differential equations) and impenetrable to anyone else.
14.03.2007 02:31
Terrible question indeed. ODE's of this exact form can be classified neatly without any DE theory, though. Let $u$ and $v$ be solutions to $y''+\alpha y=0$. Then there exists a real number number $k$ such that $u'v-uv'=k$. (Compute the derivative of $u'v-uv'$). Now given two solutions $u, v$ such that the Wronskian (let's be honest) above is not zero, it is easily shown by elementary means (use the above) that any solution of the equation is a linear combination of $u$ and $v$.
14.03.2007 04:16
Outrageous problem indeed! However, I was thinking about some particular functions, maybe : \[g(x)=\frac{f^{\prime}(x)\cos(2007x)}{2007}+f(x)\sin(2007x).\] Then, taking some derivatives stuff like that... Maybe it's not that bad after all; what do you think guys?