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Consider the initial value problem $y'' = - y$ with $y(0) = 0$, $y'(0) = 1$, $0 \leq x \le \pi$. Without solving the differential equation, show that $y$ is symmetric about $m$, where $y'(m) = 0$.

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By iterating on $y''=-y$, we see that $y$ is infinitely differentiable. Expand $y(m+x)$ around $m$: $$ y(m+x)=y(m)+y'(m)x+\frac{y''(m)}{2}x^2+\frac{y'''(m)}{3!}x^3+\ldots $$ Because $y'(m)=0$ and $y''=-y$, all the derivatives with odd orders vanish. This implies $y(m+x)=y(m-x)$.

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Thank you Kim. But how do we get the assurance that $y$ is infinitely differentiable. – user144660 Aug 25 '14 at 9:16
$y''=-y$ so $y'''=-y'$ and $y^{(4)}=-y''=y$. You can continue this indefinitely. Kim out (to bed!). – Kim Jong Un Aug 25 '14 at 9:18

The initial value problem $$ y''=-y,\quad y(m)=y_0,\,\, y'(m)=0,\qquad (\star) $$ enjoys uniqueness. Let $\psi$ be a solution and set $\varphi(x)=\psi(2m-x)$. Clearly $$ \varphi''(x)=\psi''(2m-x)=-\psi(2m-x)=-\varphi(x),\\ \varphi(m)=\psi(2m-m)=\psi(m)=y_0,\\ \varphi'(m)=-\psi(2m-m)=\psi(m)=0. $$ Thus $\varphi$ satisfies $(\star)$ as well, and hence $\varphi\equiv\psi$, due to uniqueness.

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Superb. Thank you Yiorgos. The logic is very nice. – user144660 Aug 25 '14 at 9:17

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