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I have never done a differential equation like this so please bear with me as I try to explain how I approached it.

I have $\theta(t)$ which defines the trajectory of a particle in a field.

The equation I have obtained is:

$\dot{\theta(t)}^2 = k(-cos\theta)$ where $k$ is some constant. The negative sign in front of $cos(\theta)$ is not a problem since the physical limit I obtain strictly restricts $cos(\theta)$ to have negative values such that $\dot{\theta(t)}^2$ is positive (or zero). The question asks me to find $\theta(t)$ and keep it in an indefinite integral form, and I do not even know how to begin this problem.

The attempt that I made is:

$\dot{\theta} = \sqrt{k}\sqrt{-cos(\theta)}$ which is a differential equation. so I get $\frac{d\theta}{\sqrt{-cos(\theta)}} = \sqrt{k} dt$.

However, it does not seem like I am on the right track, cause I do not know what comes next. Putting this in mathematica gives me an elliptic function. Any help would be appreciated.

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  • $\begingroup$ You can make a replacement $$t=tan(\frac{\theta }{2}) $$ $\endgroup$ – vic165 Nov 18 '19 at 4:26
  • $\begingroup$ $$ cos(\theta)=\frac{1-t^2}{1+t^2}$$ $\endgroup$ – vic165 Nov 18 '19 at 4:30
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The next step is to integrate both sides over the path. If you have an initial condition of $\theta(t_0) = \theta_0$, then you integrate both sides from that point to $\{t,\theta(t)\}$: $$ \int_{\theta_0}^{\theta(t)} \frac{d\theta'}{\sqrt{-\cos\theta'}} = \int_{t_0}^t\sqrt{k} \,dt' = \sqrt{k}(t-t_0). $$ Now, this isn't going to simplify without using special functions, and since the problem calls for a solution in terms of an indefinite integral, I think it wants us to stop here. Thus, we have an equation for the path $\theta(t)$ in implicit form: $$ \int_{\theta_0}^{\theta(t)} \frac{d\theta'}{\sqrt{-\cos\theta'}} =\sqrt{k}(t-t_0). $$

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