Let's take a Schwarzschild black hole as an example.
For a Schwarzschild black hole, we have $${\rm d}s^2=-\left(1-\frac{2GM}{r}\right){\rm d}t^2+\left(1-\frac{2GM}{r}\right)^{-1}{\rm d}r^2+r^2{\rm d}\theta^2+r^2{\rm sin}^2\theta{\rm d}\varphi^2$$So the metric tensor ##g_{\mu\nu}## has only diagonal element: $$g_{00}=-\left(1-\frac{2GM}{r}\right),\;g_{11}=\left(1-\frac{2GM}{r}\right)^{-1},\;g_{22}=r^2,\;g_{33}=r^2{\rm sin}^2\theta$$Therefore, we could calculate the connection ##\Gamma^{\mu}_{\nu\lambda}##: $$\Gamma^{0}_{01}=\Gamma^{0}_{10}=\frac{GM}{r\left(r-2GM\right)}$$ $$\Gamma^{1}_{00}=\frac{GM}{r^3}\left(r-2GM\right),\;\Gamma^{1}_{11}=-\frac{GM}{r\left(r-2GM\right)}$$ $$\Gamma^{1}_{22}=-\left(r-2GM\right),\;\Gamma^{1}_{33}=-\left(r-2GM\right){\rm sin}^2\theta$$ $$\Gamma^{2}_{21}=\Gamma^{2}_{12}=\Gamma^{3}_{31}=\Gamma^{3}_{13}=\frac{1}{r}$$ $$\Gamma^{2}_{33}=-{\rm sin}\theta{\rm cos}\theta,\;\Gamma^{3}_{23}=\Gamma^{3}_{32}={\rm cot}\theta$$
Consider the geodesics equation $$\frac{{\rm d}^2x^{\mu}}{{\rm d}p^2}+\Gamma^{\mu}_{\nu\lambda}\frac{{\rm d}x^{\nu}}{{\rm d}p}\frac{{\rm d}x^{\lambda}}{{\rm d}p}=0$$where ##p## is an affine parameter (##p=\tau## if the particle has mass ##m##, where ##\tau## is proper time), we can conclude that$$\frac{{\rm d}^2t}{{\rm d}p^2}+\frac{2GM}{r\left(r-2GM\right)}\frac{{\rm d}r}{{\rm d}p}\frac{{\rm d}t}{{\rm d}p}=0$$ $$\frac{{\rm d}^2r}{{\rm d}p^2}+\frac{GM}{r^3}\left(r-2GM\right)\left(\frac{{\rm d}t}{{\rm d}p}\right)^2-\frac{GM}{r\left(r-2GM\right)}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2-\left(r-2GM\right)\left[\left(\frac{{\rm d}\theta}{{\rm d}p}\right)^2+{\rm sin}^2\theta\left(\frac{{\rm d}\varphi}{{\rm d}p}\right)^2\right]=0$$ $$\frac{{\rm d}^2\theta}{{\rm d}p^2}+\frac{2}{r}\frac{{\rm d}r}{{\rm d}p}\frac{{\rm d}\theta}{{\rm d}p}-{\rm sin}\theta{\rm cos}\theta\left(\frac{{\rm d}\varphi}{{\rm d}p}\right)^2=0$$ $$\frac{{\rm d}^2\varphi}{{\rm d}p^2}+\frac{2}{r}\frac{{\rm d}r}{{\rm d}p}\frac{{\rm d}\varphi}{{\rm d}p}+2{\rm cot}\theta\frac{{\rm d}\theta}{{\rm d}p}\frac{{\rm d}\varphi}{{\rm d}p}=0$$
Since the gravitational field is isotropic, we can define the plane constructed by the particle's position vector ##\vec{r}## and momentum vector ##\vec{p}## when ##t=0## as the equatorial plane, which means that ##\theta=\pi/2,\;\dot{\theta}=0##, and thus we know (from the third geodesics equation) that ##\ddot{\theta}=0##.
Due to the property of Killing vector field, we can get two conserved quantities: ##E\equiv \left(1-2GM/r\right){\rm d}t/{\rm d}p={\rm constant}## and ##L\equiv r^2{\rm sin}^2\theta{\rm d}\varphi/{\rm d}p=r^2{\rm d}\varphi/{\rm d}p={\rm constant}##
Besides, we know that ##-g_{\mu\nu}u^{\mu}u^{\nu}=\epsilon##. For photons we have ##\epsilon=0##, and for particles with mass ##m##, ##\epsilon=1##. Therefore, we can conclude that $$-\left(1-\frac{2GM}{r}\right)\left(\frac{{\rm d}t}{{\rm d}p}\right)^2+\left(1-\frac{2GM}{r}\right)^{-1}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+r^2\left(\frac{{\rm d}\varphi}{{\rm d}p}\right)^2=-\epsilon$$replace ##\left(1-2GM/r\right){\rm d}t/{\rm d}p## with ##E## and ##r^2{\rm d}\varphi/{\rm d}p## with ##L##, we know that $$-E^2+\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+\left(1-\frac{2GM}{r}\right)\left(\frac{L^2}{r^2}+\epsilon\right)=0$$ which means $$\frac{1}{2}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+V\left(r\right)=\frac{1}{2}E^2$$where ##V\left(r\right)=\epsilon/2-GM\epsilon/r+L^2/\left(2r^2\right)-GML^2/r^3##
In conclusion, we have the equations of motion as follow $$\theta=\frac{\pi}{2}\\\left(1-\frac{2GM}{r}\right)\frac{{\rm d}t}{{\rm d}p}=E\\r^2\frac{{\rm d}\varphi}{{\rm d}p}=L\\\frac{1}{2}\left(\frac{{\rm d}r}{{\rm d}p}\right)^2+V\left(r\right)=\frac{1}{2}E^2$$