Solving differential Equation for the Two-Body Problem So, I'm following the derivation in D. Morin, Introduction to Classical Mechanics, of the equations for a two-body system. I understand all of it, aside from this one step. 
When he's talking about solving Lagrangian for $r(\theta)$, I don't follow the step. 
$$\tag{*}L = \frac {1}{2} m\dot{r}^{2}+ \frac{\ell^2}{2mr^2}+u(r)$$
and some how he ends with 
$$\tag{7.16}(\frac{1}{r^2} \frac{dr}{d \theta})^2 = \frac{2mE}{\ell^2}- \frac{1}{r^2} - \frac{2mu(r)}{\ell^2}.$$
This is probably a case where my lack of differential equation skills causes trouble. 
 A: I think there is a sign problem in one of your formulae, however, with your conventions, this is my derivation:
The lagrangian L is $L = \frac {1}{2} m\dot{r}^{2}+ \frac{l^2}{2mr^2}+u(r) = K - V$
where K is the kinetic energy $\frac {1}{2} m\dot{r}^{2} + \frac{l^2}{2mr^2}$ and $V$ is the potential $V(r) = - u(r)$
The constant energy is then $E = K + V = \frac {1}{2} m\dot{r}^{2} + \frac{l^2}{2mr^2} - u(r)$
Multiplying by $\frac{2m}{l^2}$, we get : 
$\frac{2m}{l^2} E = \frac{2m}{l^2}(\frac {1}{2} m\dot{r}^{2}) + \frac{1}{r^2} - \frac{2m}{l^2} u(r)$
With $l = m \dot\theta r^2$, you have $\frac{2m}{l^2}(\frac {1}{2} m\dot{r}^{2}) = \frac{2m}{(\large m \dot\theta r^2)^2}(\frac {1}{2} m\dot{r}^{2}) = (\frac{\dot r}{\dot \theta r^2})^2 = (\frac{1}{r^2} \frac{dr}{d \theta})^2$, So we have : 
$$\frac{2m}{l^2} E = (\frac{1}{r^2} \frac{dr}{d \theta})^2 + \frac{1}{r^2} - \frac{2m}{l^2} u(r)$$, that is : 
$$(\frac{1}{r^2} \frac{dr}{d \theta})^2 = \frac{2m}{l^2} E  - \frac{1}{r^2} + \frac{2m}{l^2} u(r)$$
So I do not understand the minus sign for the last term of your second formula.
