# How do we solve such differential equations?

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

$\frac {\mathrm d }{\mathrm d t}(\frac{\mathrm d\theta}{\mathrm dt})+gsin\theta=0$

How can I solve the above equation? I've never encountered anything like this before.

Thanks.

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The solution involves Jacobi's elliptic function "sn". There is a (very sketchy) description of how to integrate the equation at mai.liu.se/~halun/complex/elliptic. –  Hans Lundmark Feb 1 '12 at 15:22
you can solve a small angle approximation by assuming $\sin\theta\approx\theta$ –  yoyo Feb 1 '12 at 15:24
@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. –  Green Noob Feb 1 '12 at 15:31

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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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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Can you explain how it is in separable form? I'm a beginner so please bear with me.. –  Green Noob Feb 1 '12 at 15:45
$\int v\,dv=-g\cdot \int \sin \theta \,d\theta$ –  pedja Feb 1 '12 at 15:48
Thanks. Seems obvious now that you have shown it. –  Green Noob Feb 1 '12 at 15:54
then you have to integrate something like $$\frac{\theta'}{\sqrt{2g\cos\theta+C}}?$$ –  yoyo Feb 1 '12 at 16:07
@yoyo I simplified a step further & got this equation : $\frac{d\theta}{\sqrt{2C+2gcos\theta}}=dt$ I'm not sure I can integrate this. –  Green Noob Feb 1 '12 at 16:09
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