I am working through question 2 of the 2016 Part III exam on General Relativity
We are asked to consider the Vaidya metric:
$$d s^2 = g_{\mu\nu}\,d x^\mu \,d x^\nu = - \left( 1 - \frac{2M(v)}{r}\right) \, d v^2 + 2 \, d v \, d r + r^2 \, d\theta^2 + r^2 \sin^2\theta \, d\phi^2$$
and calculate the Christoffel symbols by determining the geodesic equation from the Lagrangian $L = -g_{\mu\nu}\frac{dx^\mu}{d \tau}\frac{dx^\nu}{d \tau}$. I found the Lagrangian to be:
$$L = \biggl( 1 - \frac{2M(v)}{r}\biggr) \, \dot{v}^2 - 2 \dot{v}\dot{r} -r^2\dot{\theta}^2 -r^2 \sin^2\theta \, \dot{\phi}^2$$
where a dot denotes $d/d \tau$. Then, I obtained the following for the $r$ and $v$ components of the geodesic equation:
$$\ddot{r} - \biggl( 1 - \frac{2M(v)}{r}\biggr) \, \ddot{v} + \biggl( \frac{M'(v)}{r}\biggr) \, \dot{v}^2 - \biggl( \frac{2M(v)}{r^2}\biggr) \, \dot{v}\dot{r}=0$$
$$\ddot{v} + \biggl( \frac{M(v)}{r^2}\biggr) \, \dot{v}^2 - r\dot{\theta}^2 - r \sin^2\theta \, \dot{\phi}^2 = 0$$
By substituting the second equation into the first, we obtain:
$$\ddot{r} + \biggl( \biggl( 1- \frac{2M(v)}{r}\biggr) \biggl( \frac{M(v)}{r^2} \biggr) + \frac{M'(v)}{r}\biggr) \, \dot{v}^2 + (2M(v)-r)\, \dot{\theta}^2 + (2M(v)-r) \sin^2\theta \, \dot{\phi}^2 - \biggl( \frac{2M(v)}{r^2}\biggr) \, \dot{v}\dot{r} = 0$$
which, together with the equations for $\theta$ and $\phi$, give us the Christoffel symbols. After that, we are asked to consider the above two equations for the special case in which $\theta$ and $\phi$ are constants.
$$\ddot{v} + \biggl( \frac{M(v)}{r^2}\biggr) \, \dot{v}^2 = 0$$
$$\ddot{r} + \biggl( \biggl( 1- \frac{2M(v)}{r}\biggr) \biggl( \frac{M(v)}{r^2} \biggr) + \frac{M'(v)}{r}\biggr) \, \dot{v}^2 - \biggl( \frac{2M(v)}{r^2}\biggr) \, \dot{v}\dot{r} = 0$$
At this point we are asked to "Show that $(1)$ curves satisfying $v = \text{const}$, are radial null geodesics and $(2)$ that curves satisfying $$ \frac{d r}{d v} = \frac{1}{2} \biggl( 1 - \frac{2M(v)}{r}\biggr) \tag{$\dagger$}$$ are radial null geodesics."
Part (1) follows immediately from the equation for $\ddot{r}$, but I am having difficulty in doing part (2). If the $\frac{M'(v)}{r}$ term were not there, then by dividing the $\ddot{r}$ equation by $\dot{v}^2$ and using ($\dagger$) we get the result fairly quickly. That term came from the Euler-Lagrange equation for the $v$ component:
\begin{align} 0 & = \frac{d}{d\tau}\left(\frac{\partial L}{\partial \dot{v}}\right) - \frac{\partial L}{\partial v}\\ & = \frac{d}{d \tau} \biggl( 2 \, \biggl( 1 - \frac{2M(v)}{r}\biggr) \, \dot{v} - 2 \dot{r} \biggr) + \frac{2M'(v)}{r} \, \dot{v}^2 \\ & = \, \biggl( 1 - \frac{2M(v)}{r}\biggr) \, \ddot{v} - \biggl( \frac{2M'(v)\dot{v}r-2M(v)\dot{r}}{r^2}\biggr) \, \dot{v} - \ddot{r} + \frac{M'(v)}{r} \, \dot{v}^2 \\ & = \, \biggl( 1 - \frac{2M(v)}{r}\biggr) \, \ddot{v} - \biggl( \frac{M'(v)}{r}\biggr) \, \dot{v}^2+ \biggl( \frac{2M(v)}{r^2}\biggr) \, \dot{v}\dot{r} - \ddot{r} \\ & = \, \ddot{r} - \biggl( 1 - \frac{2M(v)}{r}\biggr) \, \ddot{v} + \biggl( \frac{M'(v)}{r}\biggr) \, \dot{v}^2 - \biggl( \frac{2M(v)}{r^2}\biggr) \, \dot{v}\dot{r} \end{align}
I have a feeling that it should have cancelled somewhere, but I can't find where my calculations went wrong. Or maybe I need to use $(\dagger)$ to show $M'(v)=0$ but I don't think that's the case. Also, I haven't used the $\ddot{v}$ equation for anything in this part...
Any suggestions would be much appreciated.