Perihelion Advance in curve space I've been doing some work that involves particle's motion subjected to central forces. I was trying to compute the perihelion advance of the particle's path. For that I've been following this reference, where found the equation of motion, I considered a small perturbation to circular orbit. The problem was that instead of the typical harmonic oscillator ODE I ended up with an equation of the form:
$$\ddot{x}+A~\dot{x}^2 + B~x=0,$$
where $A$ and $B$ are constants. Can we retrieve any information from this equation without actually solving it, I mean, in the sense that we can find an expression for the apsidal angle if the equation was of an harmonic oscillator?
Note: In case someone is wondering why the appearance of the $\dot{x}^2$ term, it is because the considered space is curved.
 A: Your question asks about perihelion advance in curved space but I don't see how your reference link is related to that?
Assuming that you really do want to calculate perihelion precession in a curved space-time ('-helion' refers to the sun, periastron is the distance of closest approach to a general star), you might start-off the calculation using the Schwarzschild metric. This could provide some guidelines for the details of your approach and perhaps give you additional physical insight into the significance of your 2nd order equation. 
Assuming that this is a helpful approach to take, a straightforward way to do the calculation is to use a standard technique of nonlinear analysis: the phase-plane approach (see this Ref).  I mention this because the same approach can be applied to your equation as well, in a straightforward way.
Keep in mind that in the usual textbook calculation of perihelion precession that there are generally 2 approaches taken: (a) approximate an elliptic integral, or (b) a perturbative solution to the general relativistic equations.  Personally, I think the phase-plane approach is simpler and more intuitive, the calculation is shown below. 
Starting from the Schwarzschild metric we can derive a 2nd order differential equation describing the orbit (in dimensionless form):
$${{{d^{\,2}}u} \over {d{\varphi ^2}}} + u = \sigma  + {\textstyle{3 \over 2}}{u^{\,2}}$$
where $u = {r_s}/r$, $r_s$ is the Schwarzschild radius in MKS units, G is Newton’s gravitational constant, c is the speed of light, and $\varphi$ is defined in the equatorial plane:
$${r_s} = 2MG/{c^2}.$$
The dimensionless parameter $\sigma$ is given by
$$\sigma  = {\textstyle{1 \over 2}}{\left( {{m}{\kern 1pt} c{\kern 1pt} {r_s}/J} \right)^2} = 2{\left( {GM{m}/cJ} \right)^2} = {\textstyle{1 \over 2}}{\left( {{r_s}/J} \right)^2}$$
where $J$ is the angular momentum (a const. of the motion) and the RHS is expressed  using the so-called “geometrized” system of units. i.e., $G = c = 1, {r_s} = 2M$ with ${m<<M}$ taken as unity for comparison to standard results. 
But note that there is no damping term as in your equation.
To get the value for precession using the phase-plane approach, convert the above 2nd order equation to 2 first order equations by defining new variables: $x=u$ and $y=du/d\varphi$.  The equation is then equivalent to (primes will denote derivatives with respect to $\varphi$):
$$\eqalign{
  & x' = f(x,y) = y  \cr 
  & y' = g(x,y) = {\textstyle{3 \over 2}}{x^2} - x + \sigma  \cr} $$
To find the fixed points of the system above, i.e. equilibrium points of the solution, we solve simultaneously: $x'=y'=0$, for $x$ and $y$.  Therefore, the fixed points are given by
$$\vec x_1^* = \left( {{\textstyle{{1 + \sqrt {1 - 6\sigma } } \over 3}},0} \right)\,\,;\,\,\,\,\vec x_2^* = \left( {{\textstyle{{1 - \sqrt {1 - 6\sigma } } \over 3}},0} \right)$$
A classification of the fixed points is given by a linear stability analysis (see e.g., Strogatz) by series expanding the system about an arbitrary fixed point in the small parameters: $\delta x = x - x^*$ and $\delta y = y - y^*$.  Dropping second order terms, the equations are thus expressed in matrix form:
$$\left( {\matrix{
   {\delta x'}  \cr 
   {\delta y'}  \cr 
 } } \right) \approx {\left. {\left( {\matrix{
   {{\partial _x}f} & {{\partial _y}f}  \cr 
   {{\partial _x}g} & {{\partial _y}g}  \cr 
 } } \right)} \right|_{\vec x = \vec x*}}\left( {\matrix{
   {\delta x}  \cr 
   {\delta y}  \cr 
 } } \right) = {\left. {\left( {\matrix{
   0 & 1  \cr 
   {3x - 1} & 0  \cr 
 } } \right)} \right|_{\vec x = \vec x*}}\left( {\matrix{
   {\delta x}  \cr 
   {\delta y}  \cr 
 } } \right) \equiv {\left. A \right|_{\vec x = \vec x*}}\delta \vec x$$
In summary, $\vec x_1^*$ is an unstable "saddle" node while $\vec x_2^*$ is a "center" node.  The Schwarzschild linear stability phase-plane diagram, completely summarizing the dynamics is shown below:

To finish the perihelion calculation we simply solve the system about $\vec x_2^*$, which we write as:
$$\delta x' = \delta y\,\,,\,\,\delta y' =  - {\omega ^2}\delta x\,\,\,;\,\,\omega  = {(1 - 6\sigma )^{1/4}}$$
The solutions are of course "centers" corresponding to precessing elliptical orbits 
$$\eqalign{
  & \delta x(\varphi ) = A\cos \omega \varphi  + B\sin \omega \varphi   \cr 
  & \delta y(\varphi ) =  - \,\,\omega A\sin \omega \varphi  + \omega B\cos \omega \varphi \,, \cr} $$
with A and B arbitrary constants (these are different from your A and B).  Choosing initial conditions at the position of perihelion, $\delta x(0) = {u_0};\,\,\delta y(0) = du(0)/d\varphi  = 0$, we get
$$\eqalign{
  & \delta x(\varphi ) = u(\varphi ) = {u_0}\cos \omega \varphi   \cr 
  & \delta y(\varphi ) = u'(\varphi ) =  - \omega \,{u_0}\sin \omega \varphi \,, \cr} $$
giving a typical “center” solution about the fixed point $\vec x_2^*$.
In “physical” space the orbit of $m$ about $M$ does not close.  However, the phase-plane trajectory must close after a single orbit since the system is conservative (ignoring radiative effects).  Therefore, the period of a single orbit, $\Phi$, is defined from the period of the phase space trajectory:  $\omega \Phi  = 2\pi $.  
Solving for $\Phi$, and then substituting for $\omega$ in the limit of small $\sigma $ gives the result:
$$\Phi  = 2\pi {\omega ^{ - 1}} \approx 2\pi  + 3\pi \sigma $$
The Newtonian calculation gives only the first term, $\Phi  = 2\pi$, as expected. The Schwarzschild correction is therefore 
$$\Delta \varphi  = 3\pi \sigma  = 6\pi {\left( {GM{m}/cJ} \right)^2}$$
which is the usual value (in MKS units).
So in summary, you could carry through an analogous procedure for your differential equation, and then attempt to identify any corrections to $\Phi$.  Hopefully these notes have clarified how you can work out those details in your case.
Also of interest, and question for you, what form of "metric" could your differential equation correspond to?
