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I encountered the following differential equation when I tried to derive the equation of motion of a simple pendulum:

$\frac{\mathrm d^2 \theta}{\mathrm dt^2}+g\sin\theta=0$

How can I solve the above equation?

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    @yoyo Can't it be done without the small angle approximation? I wanted to derive a general equation for all cases. Anyways I don't know how to solve even with the approximation, so it'll be helpful even if you explain that method.2012-02-01

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Use substitution : \theta' =v ,therefore we have that :

\theta''=\frac{dv}{dt}\cdot \frac{dt}{d\theta}\cdot \frac{d\theta}{dt} \Rightarrow \theta''=\frac{dv}{d\theta}\cdot v \Rightarrow \theta''=v'\cdot v

where $v$ is function in terms of variable $\theta$ .So differential equation becomes :

v' \cdot v +g \cdot \sin \theta=0

which is separable differential equation .

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    @GreenNoob: Writing $\cos\theta=1-2\sin^2(\theta/2)$ reveals that it's an [elliptic integral](http://en.wikipedia.org/wiki/Elliptic_integral), so it can't be done with elementary functions. When you integrate, $t$ will be an elliptic integral of $\theta$, and inverting that relation gives $\theta$ as an [elliptic function](http://en.wikipedia.org/wiki/Jacobi_elliptic_functions) of $t$. (Elliptic functions are by definition inverses of elliptic integrals, similar to how one can define the sine function as the inverse of the arcsin integral $\int_0^x \frac{ds}{\sqrt{1-s^2}}$.)2012-02-02
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Start with $ \frac{1}{2}\frac{\mathrm{d}\dot{\theta}^{2}}{\mathrm{d}\theta} = \dot{\theta}\frac{\mathrm{d}\dot{\theta}}{\mathrm{d}\theta} = \frac{\mathrm{d}\theta}{\mathrm{d}t}\frac{\mathrm{d}\dot{\theta}}{\mathrm{d}\theta} = \frac{\mathrm{d}\dot{\theta}}{\mathrm{d}t} = \ddot{\theta} $ Then your equation becomes $ \frac{1}{2}\frac{\mathrm{d}\dot{\theta}^{2}}{\mathrm{d}\theta} = -\frac{g}{\ell}\sin(\theta) $ or $ \mathrm{d}\dot{\theta}^{2} = -\frac{2g}{\ell}\mathrm{d}\sin(\theta) \implies \dot{\theta}^{2} = \frac{2g}{\ell}\cos(\theta) + c_{1} $ It's a bit easier if we assume initial conditions, say $\dot{\theta}(t_{0}) = \dot{\theta}_{0}$ and $\theta(t_{0}) = \theta_{0}$, so that $ \dot{\theta}^{2} = \frac{2g}{\ell}\left[\cos(\theta)- \cos(\theta_{0}) + \frac{\ell\dot{\theta}_{0}^{2}}{2g} \right] $ Then $ \frac{\mathrm{d}\theta}{\mathrm{d}t} = \sqrt{ \frac{2g}{\ell}}\sqrt{\cos(\theta) - \cos(\theta_{0}) + \frac{\ell\dot{\theta}_{0}^{2}}{2g} } $ so that $ \mathrm{d}t = \sqrt{\frac{\ell}{2g}}\frac{\mathrm{d}\theta}{ \sqrt{\cos(\theta) - \cos(\theta_{0}) + \frac{\ell\dot{\theta}_{0}^{2}}{2g}}} $ or $ t_{f} - t_{0} = \sqrt{\frac{\ell}{2g}}\int_{\theta_{0}}^{\theta_{f}}\frac{\mathrm{d}\theta}{ \sqrt{\cos(\theta) - \cos(\theta_{0}) + \frac{\ell\dot{\theta}_{0}^{2}}{2g}}} $ This equation is of the form $t = f(\theta)$. Your solution is given by $\theta = f^{-1}(t)$. That's about as much as you need to know, since it's more efficient to just solve the original equation numerically.

If you really need a closed form for $f$, Mathematica will give you one, in terms of the function EllipticF.

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replacing $\sin\theta$ by $\theta$ (physically assuming small angle deflection) gives you a homogeneous second order linear differential equation with constant coefficients, whose general solution can be found in most introductory diff eq texts (or a google search). this new equation represents a simple harmonic oscillator (acceleration proportional to displacement, like a spring force). \theta''+g\theta=0 has solutions $A\cos(\sqrt{g}t)+B\sin(\sqrt{g}t)$. so, for example, if the initial displacement is $\theta_0$ and initial angular velocity is $0$ then the solution is $ \theta_0\cos(\sqrt{g}t) $

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    This doesn't exactly answer the question since this is just for an approximation to the original differential equation.2014-06-08