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Consider the differential equation $$\frac{d^{2}y}{dx^{2}}+y=0$$ with initial conditions $y(0)=0$ and $y'(0)=1$. The solution is well known - $y=\sin(x)$. I know how to derive this solution, since the given equation is a linear differential equation with constant coefficients, and characteristic equation $z^{2}+1=0$.

I also know that this identity, combined with the initial conditions, allows us to compute $y^{(n)}(0)$ and thus the Maclaurin series of $y$, which coincides with the Maclaurin series of $\sin(x)$. Neither of these proofs appear to use any property of $\sin(x)$ other than its oscillating derivatives.

Does there exist a proof of the solution to this equation which uses some other properties to $\sin(x)$? If not, is there a way of visualising it, considering the connection of $\sin$ to the unit circle?

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There is a way of visualizing it: view the differential equation as one involving a vector of two coordinates (one of them will be sine) - it is then geometrically intuitive that (considering the trajectory around the unit circle with constant unit speed) the derivative is the tangent vector and the second derivative the normal vector. –  anon Dec 19 '12 at 11:32

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The geometry is the following. Suppose you have a pair of functions $c(t), s(t)$ which smoothly parameterize the unit circle at unit speed. The first requirement means that $c^2 + s^2 = 1$, and the second requirement means that $c'^2 + s'^2 = 1$. Differentiating the first requirement gives $2c c' + 2s s' = 0$, hence $(c, s)$ and $(c', s')$ are orthogonal unit vectors. By continuity the angle between them is constant, so WLOG

$$c' = -s, s' = c$$

from which it follows that $s'' = -s$ and $c'' = -c$.

There is also a physical interpretation. $\frac{d^2 y}{dt^2} = -y$ describes the motion of a classical particle on the real line of mass $1$ under the influence of the potential $V(y) = \frac{1}{2} y^2$, so by conservation of energy the quantity $\frac{1}{2} y^2 + \frac{1}{2} y'^2$ is constant (so this gives a converse to the above). More generally see Hamiltonian mechanics.

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Thank you very much for this answer! –  Daniel Littlewood Dec 19 '12 at 12:28

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