# Solving the following differential equation

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.

• You can make a replacement $$t=tan(\frac{\theta }{2})$$ – vic165 Nov 18 '19 at 4:26
• $$cos(\theta)=\frac{1-t^2}{1+t^2}$$ – vic165 Nov 18 '19 at 4:30

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: $$\int_{\theta_0}^{\theta(t)} \frac{d\theta'}{\sqrt{-\cos\theta'}} =\sqrt{k}(t-t_0).$$