Geodesic equations I am having trouble understanding how the following statement (taken from some old notes) is true:

For a 2 dimensional space such that $$ds^2=\frac{1}{u^2}(-du^2+dv^2)$$
  the timelike geodesics are given by $$u^2=v^2+av+b$$ where $a,b$ are constants.

When I see "geodesics" I jump to the Euler-Lagrange equations. They give me
$$\frac{d}{d\lambda}(-2\frac{\dot u}{u^2})=(-\dot u^2+\dot v^2)(-\frac{2}{u^3})\\
\implies \frac{\ddot u}{u^2}-2\frac{\dot u^2}{u^3}=\frac{1}{u^3}(-\dot u^2+\dot v^2)\\
\implies u\ddot u-\dot u^2-\dot v^2=0$$
and 
$$\frac{d}{d\lambda}(2\frac{\dot v}{u^2})=0\\
\implies \dot v=cu^2$$
where $c$ is some constant.
Timelike implies $$\dot x^a\dot x_a=-1$$ where I have adopted the $(-+++)$ signature.
I can't for the life of me see how the statement results from these. Would someone mind explaining? Thanks.
 A: I prefer to use Killing vectors and conservation laws to solve stuff like this, so let's analyze the problem using Killing vectors, and see if the results agree with your Euler-Lagrange equations.
Notice that the metric is invariant under translations of $v$.  The associated killing vector is $\partial_v$ which in turn gives the following conserved quantity:
$$
  c_v = \dot x\cdot \partial_v = \frac{\dot v}{u^2}
$$
This agrees precisely with your second Euler-Lagrange equation; so far so good.  The timelike condition $\dot x\cdot\dot x = -1$ can be written in components as
$$
  \frac{-\dot u^2 + \dot v^2}{u^2} = -1
$$
Using the above conservation equation to eliminate $\dot v$ then gives a first order differential equation for $\dot u$
$$
  \frac{-\dot u^2+c_v^2u^4}{u^2}=-1
$$
which simplifies to
$$
  \dot u^2 = c_v^2u^4 + u^2
$$
This is a first order, separable differential equation that can be solved by separation of variables and integration.  Once you solve this for $u$, you can plug the solution back into the conservation equation $c_v = \dot v/u^2$ and solve this equation by integration as well.  This yields the general solution to the system of differential equations, and then you can relate $u$ and $v$ in the way stated in the quote.
Warning; there may be simpler ways of showing what you want to show.
A: Method 1: Implicit differentiation without explicitly solving ODEs:
$$\frac{d(u^2)}{dv}~=~\frac{1}{\dot{v}} \frac{d(u^2)}{dt}~=~\frac{2u\dot{u}}{\dot{v}}~=~\frac{2\dot{u}}{cu} $$ 
$$\Downarrow $$
$$\frac{d^2(u^2)}{dv^2}~=~\frac{1}{\dot{v}} \frac{d}{dt}\left(\frac{2\dot{u}}{cu} \right)~=~\frac{2}{\dot{v}} \frac{\ddot{u}u-\dot{u}^2}{cu^2}=\frac{2}{\dot{v}} \frac{\dot{v}^2}{cu^2} ~=~2,$$
which in turn implies OP's sought-for equation. Above we have only used the two Euler-Lagrange equations $\ddot{u}u=\dot{u}^2+\dot{v}^2$ and $\dot{v}=cu^2$.
Method 2: Explicitly solving ODEs:
Rescale the variables as
$$ U~:=~cu, \qquad V~:=~cv. $$
Then the two equations $\dot u^2 = c^2u^4 + u^2$ and $\dot{v}=cu^2$ become
$$ \dot{U}^2 ~=~U^4+U^2, \qquad \dot{V}~=~U^2 ,$$
with full solution 
$$U(t)~=~\pm {\rm csch}(t-t_0),  \qquad V(t)~=~\coth(t-t_0)+V_0. $$
OP's sought-for equation now follows from
$$(V-V_0)^2 ~=~ U^2 +1.$$
