I'm trying to calculate the Killing vectors associated to a 2D torus, parametrised as $ds^2=a^2d\theta^2+(b+a\sin\theta)^2d\phi^2$. From here I got the metric and found out that the only connection coefficients that don't vanish are:
$$ \Gamma^\theta{}_{\phi\phi}=-\frac{(b+a\sin\theta)\cos\theta}{a} \hspace{1cm}\Gamma^\phi{}_{\theta\phi}=\Gamma^\phi{}_{\phi\theta}=\frac{a\cos\theta}{b+a\sin{\theta}}$$
From here, and given the Killing condition: $\nabla_\alpha \xi_\beta + \nabla_\beta \xi_\alpha =\partial_\alpha \xi_\beta + \partial_\beta \xi_\alpha-2\Gamma^{\lambda}_{\beta\alpha}\xi_\lambda=0$, one can get the three following equations for the killing vectors:
$$\partial_\phi \xi_\phi = \Gamma^\theta{}_{\phi\phi}\xi_\theta=-\frac{(b+a\sin\theta)\cos\theta}{a}\xi_\theta \hspace{2cm} (1)$$ $$\partial_\phi \xi_\theta + \partial_\theta \xi_\phi = \frac{2a\cos\theta}{b+a\sin{\theta}}\xi_\phi \hspace{2cm} (2)$$ $$ 2\partial_\theta \xi_\theta=0 \hspace{2cm} (3)$$
From (3) it's easy to see that $\xi_\theta=f(\phi)$, so that (1) can be solved as:
$$\xi_\phi = -\frac{(b+a\sin\theta)\cos\theta}{a}F(\phi)+g(\theta),$$ with $F'(\phi)=f(\phi)$. This way, one can substitute in (2) and get the following differential equation:
$$f'(\phi)+[-\cos(2\theta) + \frac{b}{a}\sin\theta]F(\phi)=- \frac{2a\cos\theta}{b+a\sin\theta}g(\theta)-g'(\theta)$$
But I can't see how to solve this (could be that my differential equation solving skills aren't pretty much on point though... If only I could separate variables), and wouldn't like to spend too much time on it since it could also be that I made some mistake in the previous calculations, so if anyone sees something wrong (or that I'm on the good way), any feedback will be much appreciated!
Look at your "final differential equation". Take derivative w.r.t. $\phi$. The RHS vanishes. The LHS becomes
$$ f''(\phi) - (\cos(2\theta) - \frac{b}{a} \sin\theta ) f(\phi) = 0.$$
Taking now the $\theta$ derivative, you find $$ \left( 2 \sin(2 \theta) + \frac{b}{a} \cos(\theta) \right) f(\phi) = 0 $$
This equation is required to hold everywhere, and specifically when $\theta = 0$, which requires $f \equiv 0$. (In the case where $b = 0$ you no longer have a torus, and you do have other solutions!)
Plugging this into (1) you find that $\xi_\phi = h(\theta)$. And plugging this into (2) you find that
$$ h' = \frac{2a \cos\theta}{b + a\sin\theta} h $$
which is a separable ODE which you easily solve to find
$$ h(\theta) = C (b+a\sin\theta)^2 $$
So in conclusion:
the only Killing vector field of the 2D torus is the obvious one, corresponding to $\partial_\phi$.