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The Laplacian is differential operator defined by $$\Delta U=\sum^n_{i=1} \frac{\partial^2 u}{\partial x^2 _i}$$

Find at least one solution to $$-\Delta u(x)=u(x)$$

How do I solve this equation? The solution says $$u(x) = c1 cos x + c2 sin x$$ but I have no idea where this comes from.

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Since $u(x)$ appears to only be a function of one coordinate (although the notation is a bit confusing without context), the other terms in the Laplacian ($\partial^2 u/\partial y^2,\partial^2 u/\partial z^2$, &c.) are zero, so the equation reduces to $$ -u''(x) = u(x), $$ which you probably know is solved by linear combinations of $\sin{x}$ and $\cos{x}$.

Chappers
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  • I did not know that it was solved by a combination of sin x and cos x, why is that? –  Mar 31 '15 at 15:40
  • $\sin'{x}=\cos{x}$, and $\cos'{x}=-\sin{x}$, so $\sin''{x}=-\sin{x}$ and $\cos''{x}=-\cos{x}$. – Chappers Mar 31 '15 at 15:43