How do I squeeze a $\theta(t)$ and $\varphi(t)$ out of this?

Question

A ball attached to a fixed-length massless rod swings about under gravity. Mathematically:

$$L=T-U=\frac{MR^2}{2}(\sin^2(\theta)\dot{\varphi}^2+\dot{\theta}^2)+MgR \cos(\theta)$$

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)=\frac{\partial L}{\partial \theta}$$ $$MR^2 \ddot{\theta}=MR^2\sin(\theta)\cos(\theta)\dot{\varphi}^2-MgR \sin(\theta)$$ $$MR^2 \ddot{\theta}=\frac{MR^2}{2}\sin(2\theta)\dot{\varphi}^2-MgR \sin(\theta)\tag{1}$$ $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi}}\right)=\frac{\partial L}{\partial \varphi}$$ $$\frac{d}{dt}(MR^2 \sin^2(\theta) \dot{\varphi})=0$$ $$MR^2( \sin^2(\theta) \ddot{\varphi}+2\sin(\theta)\cos(\theta) \dot{\varphi}\dot{\theta})=0$$ $$MR^2( \sin^2(\theta) \ddot{\varphi}+\sin(2\theta) \dot{\varphi}\dot{\theta})=0\tag{2}.$$

Any ideas as to the domestication of these equations? Any approximative tricks?

Edit: I forget that it's useful if I post some of my own insights to aid answerers

1. $\sin(\theta)=\theta+o(\theta^3)$, so approximate $\sin(\theta) \approx \theta$
2. For cosine, it's not so easy, because $cos(\theta)=1+o(\theta^2)$ doesn't hold for as long, so perhaps $cos(\theta) \approx 1-\frac{1}{2}\theta^2$ could work, but there's enough trouble already with the equations being nonlinear before adding a squared term.

Answer 1

There is no closed form solution for this, unless you assume $\theta \ll1$. In that case you get from the second equation that: $$\theta\dot{\phi}=C$$ This really doesn't help you unless you make some further assumptions, namely that either $\dot{\phi}$ or $\dot{\theta}$ are equal to 0. I've seen solutions that try to perturb the system around one of these assumptions and then use a Taylor series to get an approximate solution.

As one of my physics professors told us:

All problems that can be solved exactly will be taught to you in your undergraduate degree. From then on, it's all approximations.