Starting with the Lagrangian $\mathcal{L} = g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu$, derive $4$ equations which govern geodesics in a Schwarzschild spacetime.
My lecturer said that
If your Euler-Lagrange equation reads $$\large \frac{\mathrm{d}(\text{blah})}{\mathrm{d}\lambda} = 0$$ then don't expand it but instead deduce that blah $=$ constant.
But I'm not sure how to process this in such a general setting. Would it be like $$\large \mathcal{L} = g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu,\quad \frac{\mathrm{d}}{\mathrm{d}\lambda}\left(\frac{\partial\mathcal{L}}{\partial\dot{x}^\mu}\right) = \frac{\partial\mathcal{L}}{\partial x^\mu}$$ But then why would the RHS be a zero? I'm in need of some assistance.
Hint: Use Noether's (first) theorem: Since the Lagrangian density ${\cal L}$ does not depend explicit on "time" $\lambda$, then the "energy" function $$ h~:=~\dot{x}^{\mu} \frac{\partial {\cal L}}{\partial \dot{x}^{\mu}} -{\cal L}~=~\ldots~=~{\cal L} $$ (which happens to be ${\cal L}$ itself!) is constant in "time" along a solution. The above holds even if ${\cal L}$ depends on $x^{\mu}$!