general relativity - Transforming an equation to the co-vector version

Ok, this question is more a result of my lack of knowledge of how to manipulate equations involving index notation rather than about physics...

I have the geodesic equation with $U^\lambda\equiv\dot{x}^\lambda$:-

$$ \dot{U^\lambda} + \Gamma^\lambda_{\mu\nu} U^\mu U^\nu $$

And I want to transform to the co-vector $U_\mu=g_{\mu\lambda}U^\lambda$.

Can I simply multiply each vector by $g_{\mu\nu}$? Like so:-

$$ g_{\mu\lambda}\dot{U^\lambda} + \Gamma^\lambda_{\mu\nu}g_{\mu\alpha}U^\alpha g_{\nu\beta}U^\beta $$

Or do I need to use $g^{\sigma\nu}g_{\nu\mu} = \delta^\sigma_\mu$ to rewrite $U_\mu=g_{\mu\lambda}U^\lambda$ and then sub it in?

Edit: Here's my attempt at the sub in method

So using $g^{\lambda\mu}g_{\mu\lambda} = \delta^\lambda_\lambda$ to rewrite $U_\mu=g_{\mu\lambda}U^\lambda$ as $U^\lambda=g^{\lambda\mu}U_\mu$. (Is this even correct?). Then differentiate:- $$ \dot{U}^\lambda=\dot{g}^{\lambda\mu}U_\mu + g^{\lambda\mu}\dot{U}_\mu $$ Can I assume that the differential of the metric wrt time is going to be zero? Obivously this is not going to be true in general since massive bodies move! But generally in simple problems would this be taken as true?