Integrating the geodesic equation to obtain Newtonian energy of test particle
I am studying General Relativity, and have come across a question that I am finding rather intractable:
In Newtonian Theory, the energy equation for a test particle in orbit around a point mass is: $$\frac{v^{2}}{2} + \frac{\ell^{2}}{2r^{2}} - \frac{GM}{r} = \mathcal{E}$$ Where $r$ is the radius, $v$ is the radial velocity, $\ell$ is the angular momentum per unit mass, $\mathcal{E}$ is the constant energy per unit mass and $-GM/r$ is the gravitational potential. For the <PERSON> solution show that the integrated geodesic equation may also be written in the form: $$\frac{v_{s}^{2}}{2} + \frac{\ell_{s}^{2}}{2r^{2}} + \Phi_{s}(r) = \mathcal{E}{s}$$ Where $v{s} = \mathrm{d}r/\mathrm{d}\tau$, and $\ell_{s}$ and $\mathcal{E}_{s}$ are constants.
I am not really sure what it means by integrating the geodesic equation. My understading is that the geodesic equation is given by:
$$\frac{\mathrm{d}^{2} x^{\lambda}}{\mathrm{d} \tau^{2}} + \Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} = 0$$
Where $\Gamma^{\lambda}_{\mu\nu}$ is the affine connection. However, this leads to a system of equations, none of which, when integrated, yield the form given in the question.
I'm sure that I am missing something fundamental and simple, but I just cannot find it in any of the references or materials that I have been given.
I thought that perhaps integrating the geodesic equation directly would be the way to go, but I do not end up with the correct solution:
$$\int\left(\frac{\mathrm{d}^{2} x^{\lambda}}{\mathrm{d} \tau^{2}} + \Gamma^{\lambda}{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}\right)\:\mathrm{d}\tau = \mathcal{E}{s}$$
We note that as integration is distributive over addition, we find:
$$\frac{\mathrm{d}x^{\lambda}}{\mathrm{d}\tau} + \int\left(\Gamma^{\lambda}{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}\right)\:\mathrm{d}\tau = \mathcal{E}{s}$$
We note that if we let $\lambda = r$, then this becomes proportional to $v_{s}$ and thus I am missing a factor of $v_{s}$.
Edit 2: After the very helpful comments below, I went back and did the question again, this time getting stuck at a different point.
So we find that in the $\lambda = t$ case, the only <PERSON> symbols $\Gamma_{\mu\nu}^{t}$ that are non-zero are when $\mu = t$, $\nu = r$ or vice versa. 394