Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 5 down vote accepted

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.