Here is the trajectory in words.
There are three zones of interest:
Zone 1: x₀>1
Zone 2: -1<x₀<1
Zone 3: x₀<-1
Different forces apply in each zone, and x=1 and x=-1 are asymptotes. x₀ is the starting point for the particle and applying Newton’s Law F=ma, so a=F/m.
In Zone 1, there is a repulsive force only which causes the particle to accelerate away from x=1. We can write the acceleration a=k₁/(x-1) where k₁ is a constant which incorporates the mass of the particle. The acceleration decreases as the particle is pushed further and further away from its starting point x₀. The asymptote shields the particle from the attractive force.
In Zone 2, two forces act on the particle and the acceleration is -k₁/(1-x)-k₂/(1+x), where k₂ is the constant relating to the attractive force. The net force drives the particle towards x=-1.
In Zone 3, the attractive force is -k₂/(x+1). Since x₀<-1, x₀+1<0 which makes the acceleration increase as the particle moves towards x=-1. The asymptote shields the particle from the repulsive force.
The acceleration becomes infinite x→-1 or x→1 from either side, hence the three zones and the asymptotes.
Now the mathematical model. Let v=dx/dt. then acceleration a=dv/dt=d2x/dt2 and dx/dv.dv/dt=dx/dt=v, dx/dv=v/(d2x/dt2). So (d2x/dt2)dx=vdv. This is "integration ready" form.
Applying this to Zone 1, (k1/(x-1))dx=vdv. Integrating we get: k1ln(x-1)=v2/2+C, where C is a constant of integration. When x=x0 let v=0, meaning that initially the particle is at rest when x=x0.
C=k1ln(x0-1) and k1ln(x-1)=v2/2+k1ln(x0-1), (x-1)/(x0-1)=e^(v2/(2k1)).
Alternatively, v2=2k1ln[(x-1)/(x0-1)]. Both x and x0 are greater than 1 for this equation to apply.
Applying the equation to Zone 3, (-k2/(x+1))dx=vdv. Integrating we get: -k2ln|x+1|=v2/2+C, where C is a constant of integration. When x=x0 let v=0, meaning that initially the particle is at rest when x=x0. Remember that x0<-1 for this zone.
C=-k2ln|x0+1| and -k2ln|x+1|=v2/2-k2ln|x0+1|, (x0+1)/(x+1)=e^(v2/(2k2)). (x0+1 and x+1 are both negative.)
v2=2k2ln[(x0+1)/(x+1)].
Now to apply the equation to Zone 2:
First, simplify the formula -k1/(1-x)-k2/(1+x)=-(k1+k1x+k2-k2x)/(1-x2)=(k1+k2+k1x-k2x)/(1-x2).
(d2x/dt2)dx=vdv becomes ((k1+k2+k1x-k2x)/(1-x2))dx=-vdv.
This simplifies:
[k1/(1-x)+k2/(1+x)]dx=-vdv and integrating:
-k1ln|1-x|+k2ln|1+x|=C-v2/2, (1+x)k2/(1-x)k1=e^(C-v2/2)=(eC)(e^-(v2/2)).
If v=0 when x=x0, eC=(1+x0)k2/(1-x0)k1, [(1+x)k2/(1-x)k1][(1-x0)k1/(1+x0)k2]=e^-(v2/2).