Is there any analytic method for solving $\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$ Hi i am trying to solve the following two equations:
$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$
$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$
where $\dot{\theta}=\frac{d\theta}{dt}$ and $\dot{\phi}=\frac{d\phi}{dt}$ and $\ddot{\theta}=\frac{d^2\theta}{dt^2}$ and $\ddot{\phi}=\frac{d^2\phi}{dt^2}$. I am trying to solve this analytically and looked up the standard techniques but couldn't find how solve these. Basically i want $\phi$ in terms of $\theta$ that  is $\phi(\theta)$. Is there any way to do this?
 A: $$\frac{d^2\phi}{dt^2} + 2\cot(\theta)\frac{d\theta}{dt}\frac{d\phi}{dt} =0$$
This separable ODE is directly integrable.
$$\ln(\frac{d\phi}{dt})+2\ln(\sin(\theta))=\text{constant}$$
$$\sin^2(\theta)\frac{d\phi}{dt}=c_1\quad\implies\quad \frac{d\phi}{dt}=\frac{c_1}{\sin^2(\theta)}$$
Second equation :
$$\frac{d^2\theta}{dt^2}-\sin(\theta)\cos(\theta)\left(\frac{d\phi}{dt}\right)^2=0$$
$$\frac{d^2\theta}{dt^2}-\sin(\theta)\cos(\theta)\left(\frac{c_1}{\sin^2(\theta)}\right)^2=0$$
$$2\frac{d\theta}{dt}\frac{d^2\theta}{dt^2}-2(c_1)^2\frac{\cos(\theta)}{\sin^3(\theta)}\frac{d\theta}{dt}=0$$
$$\left(\frac{d\theta}{dt}\right)^2+(c_1)^2\frac{1}{\sin^2(\theta)}=(c_2)^2$$
$$\frac{d\theta}{dt}=\pm\sqrt{(c_2)^2-(c_1)^2\frac{1}{\sin^2(\theta)}}$$
$$t=\int \frac{\sin(\theta)}{\pm\sqrt{(c_2)^2\sin^2(\theta)-(c_1)^2 }}d\theta$$
$$t=\pm \sin^{-1}\left(\frac{c_2}{(c_1)^2+(c_2)^2}\cos(\theta) \right)+c_3$$
$$\frac{c_2}{(c_1)^2+(c_2)^2}\cos(\theta)=\pm\sin(t-c_3)$$
$$\theta(t)=\cos^{-1}\left(\pm\frac{(c_1)^2+(c_2)^2}{c_2}\sin(t-c_3) \right)$$
A: I think your system can be reduced to a Bernoulli Differential Equation: $y'(x)+p(x)\,y(x) = q(x) \, [y(x)]^n$.
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Rewrite $\dot{\phi}=\phi'(t)$ and $\ddot{\phi}=\phi''(t)$ by applying the chain rule for $\phi(\theta(t))$ and define $u(\theta)=\phi'(\theta)$ to get
$$u'(\theta) + 2 \cot(\theta) \, u(\theta) + \sin(\theta) \cos(\theta) \, [u(\theta)]^3 = 0 $$
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I will double-check this partial solution and write it down here step-by-step later.
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EDIT:
Considering the chain $\phi - \theta - t$, it's possible to write:
$$\frac{d\phi}{dt} = \frac{d\phi}{d\theta} \frac{d\theta}{dt}$$
$$\frac{d}{dt} \left[\frac{d\phi}{d\theta}\right] = \frac{d^2\phi}{dtd\theta}  = \frac{d^2\phi}{d\theta^2} \frac{d\theta}{dt}$$
$$\frac{d^2\phi}{dt^2} = \frac{d}{dt} \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] = \frac{d^2\phi}{dtd\theta} \, \frac{d\theta}{dt} + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} = \frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2}$$
Returning those results to your original system:
$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$
$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$
$$$$
$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} + 2\cot{\theta} \, \frac{d\theta}{dt} \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] =0$$
$$\frac{d^2\theta}{dt^2}- \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] ^2=0$$
$$$$
$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} + 2\cot{\theta} \, \frac{d\theta}{dt} \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] =0$$
$$\frac{d^2\theta}{dt^2} = \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^2 \left[\frac{d\theta}{dt}\right]^2$$
$$$$
$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^2 \left[\frac{d\theta}{dt}\right]^2 + 2\cot{\theta} \, \frac{d\phi}{d\theta} \, \left[\frac{d\theta}{dt}\right]^2 =0$$
$$$$
So you have a trivial equation, $[\theta'(t)]^2=0$, and
$$\frac{d^2\phi}{d\theta^2} + \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^3 + 2\cot{\theta} \, \frac{d\phi}{d\theta} =0$$
$$$$
Define $u(\theta)=\phi'(\theta)$ to get a Bernoulli Differential Equation,
$$u'(\theta) + 2 \cot(\theta) \, u(\theta) + \sin(\theta) \cos(\theta) \, [u(\theta)]^3 = 0 \text{,}$$
which presumably has analytical solutions.
A: First, we must get rid of the middle man, $t$. In other words, we will make the following observations:
$$\dfrac{d\phi}{dt}=\dot\theta\dfrac{d\phi}{d\theta}$$ and
$$\dfrac{d^2\phi}{dt^2}=\dfrac{d}{dt}\dot\theta\dfrac{d\phi}{d\theta}=\ddot\theta\phi_{\theta}+\dot\theta^2\phi_{\theta\theta}.$$
Notice now how we have $\ddot\theta$ as $\dot\phi^2\sin\theta\cos\theta$. So the first equation becomes $$\dot\theta^2\sin\theta\cos\theta\cdot\phi^3_{\theta}+\dot\theta^2\phi_{\theta\theta}+2\cot\theta\dot\theta^2\phi_{\theta}=0$$ or $$\sin\theta\cos\theta\cdot\phi^3_{\theta}+\phi_{\theta\theta}+2\cot\theta\phi_{\theta}=0$$ This is the equation for $\phi(\theta)$. If you were not too picky about the parameter $t$, this could have been solved by [other answers on this post].
Now, notice that dividing through by $\phi_\theta$, we have
$$\sin\theta\cos\theta\cdot\phi^2_{\theta}+\dfrac{\phi_{\theta\theta}}{\phi_{\theta}}+2\cot\theta=0$$ Because we have a that second term as being $$\dfrac{d}{d\theta}\log(\phi_{\theta})$$ substitute $\phi=\int e^{\beta}\mathrm{d}\theta$. Thus,
$$\cos\theta\sin\theta e^{2\beta}+\beta_{\theta}+2\cot\theta=0$$ We're almost done. Divide through by $e^{2\beta}$, and subtract the beta-like terms to obtain
$$M+Le^{-2\beta}=\dfrac{d}{d\theta}e^{-2\beta}$$ where $M=2\sin\theta\cos\theta$ and $L=4\cot\theta$. This evidently takes the form
$$y'+ay=b$$ which has the general solution $$y=e^{-\int a}(k+\int b e^{\int a})$$ for some constant $k$, and so
$$e^{-2\beta}=e^{4\int \cot\theta\mathrm{d}\theta}\left[2\int \sin\theta\cos\theta \cdot e^{-4\int\cot\theta \mathrm{d}\theta}\mathrm{d\theta}+k\right]$$ which we know that
$$\phi =\int e^{\beta}\mathrm{d\theta}=\int \dfrac{e^{-2\int \cot\theta\mathrm{d}\theta}}{\sqrt{2\int \sin\theta\cos\theta \cdot e^{-4\int\cot\theta \mathrm{d}\theta}\mathrm{d\theta}+C_0}}\mathrm{d\theta}.$$
Edit A more complete solution would be where one notes that $e^{\int \cot \theta}=|\sin \theta|$ I'm ignoring constants because they're a pain, other than the one obtained by the linear solution. Then, $$\phi =\int \dfrac{1}{\sin^2\theta\sqrt{2\int \dfrac{\cos\theta}{\sin ^3\theta}\mathrm{d\theta}+C_0}}\mathrm{d\theta}$$ And since
$$\int \dfrac{1}{\sin^3\theta}d\sin \theta=-\dfrac{1}{2\sin^2\theta}$$ we have simply that
$$\phi = \int \dfrac{1}{\sqrt{C_0\sin^4\theta-\sin^2\theta}}d\theta=\arctan\left(\sqrt{C_0\sin^2\theta-1}\right)-\dfrac{\arctan\left(\dfrac{\sqrt{C_0\sin^2\theta-1}}{\sqrt{1-C_0}}\right)}{\sqrt{1-C_0}}$$
Please comment if I made any errors.
A: Here is my attempt. Divide the first equation by $\dot\phi$, then we get
$$ \frac{d\ln\dot\phi}{dt}+2\frac{d\ln(\sin\theta)}{dt}=0.$$
We then obtain the integral of motion
$$\ln\dot\phi+2\ln\sin\theta=C.$$
Hence, we find $\dot\phi\sin^2\theta=e^C=B$.
Plugging this in the second equation, we find
$$ \ddot\theta-B^2\frac{\cos\theta}{\sin^3\theta}\dot\theta=0.$$
Multiplying this by $\dot\theta$, we find
$$\frac12\frac{d\dot\theta^2}{dt}+\frac b2\frac{d\sin^{-2}\theta}{dt}=0,$$
where $b=B^2$.
Then we get another integral of motion, a Hamiltonian $H$.
$$\frac{\dot\theta^2}2+\frac b{2\sin^2\theta}=H.$$
I believe this is right, but I don't have time to check it carefully now.
I hope this helps.
