An incorrect universe

This is my approach to this nice problem. Here is a simple diagram.

Consider only the force exerted on the right particle. It is easy to see that $F = - k_e\frac{q_1q_2}{r^3}$ as $F$ is directed towards the origin. Let $K= k_e\frac{q_1q_2}{m}$ for clarity. Therefore as $F=ma$ , $a = - \frac{K}{r^3}$ But what is the value of $r$ in our problem? By symmetry the motion of the particles must exactly mirror eachother, so it follows that $r = 2x$ . So $a = -\frac{K}{(2x)^3} = -\frac{K}{8x^3}$ It is well known that $\int a \, \mathrm{d}x = \frac{1}{2}v^2$ , so $\frac{1}{2}v^2 = \int -\frac{K}{8x^3} \, \mathrm{d}x = \frac{K}{16x^2} + c$ Substitute $\boxed{x=1, v=0}$ into the equation and solve to give $c = -\frac{K}{16}$ . Therefore $\frac{1}{2}v^2 = \frac{K}{16}\left(\frac{1}{x^2}-1\right) \\ v=-\sqrt{\frac{K}{8}\left(\frac{1-x^2}{x^2}\right)}$ Note the minus as the velocity is directed towards the origin. Now solve the differential equation: $\int - \frac{x}{\sqrt{1-x^2}} \, \mathrm{d}x = \int \sqrt{\frac{K}{8}} \, \mathrm{d}t\\ \sqrt{1-x^2} + C = \sqrt{\frac{K}{8}}t$

Substitute $\boxed{t=0,x=1}$ into the equation and solve to give $C = 0$ .

Therefore $\sqrt{1-x^2}= \sqrt{\frac{K}{8}}t$ and at $x=0$ , $\boxed{t = \sqrt{\frac{8}{K}}}$ .

Plugging in the variables gives $t = \sqrt{\frac{8}{\left(\frac{(9\times10^9) \times (1\times10^{-6}) \times (1\times10^{-6})}{1\times10^{-3}}\right)}} = \sqrt{\frac{8}{9}} = 0.9428\dots = \boxed{0.943}$

I solved this numerically via Euler's algorithm. I am curious to know how this can be done analytically.

Python 2.7

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from math import *

def dist(a,b):
    #Distance between the two vectors a and b
    x1,y1,x2,y2=a.real,a.imag,b.real,b.imag
    return sqrt((x1-x2)**2 + (y1-y2)**2)

def vect(a,b):
    #Finds the unit vector between a and b
    x1,y1,x2,y2=a.real,a.imag,b.real,b.imag
    top = (x2-x1) + (y2-y1)*1j
    return top/dist(a,b)



class charge:
    '''Represents a charge'''
    def __init__(self,q,a,v,pos):
        self.q = q
        self.a = a
        self.v = v
        self.pos = pos

    def step(self,dt):
        self.v = self.v + self.a * dt
        self.pos = self.pos + self.v * dt   


def coloumb( q1 , q2 , dt , k = 9e9, m = 1e-3):
    '''Simulates the coulomb interaction'''
    K = ((q1.q*q2.q)*(k/m))/(dist(q1.pos,q2.pos)**3)
    q1.a = K*vect(q2.pos,q1.pos)
    q2.a = K*vect(q1.pos,q2.pos)
    q1.step(dt)
    q2.step(dt)




k = 9e9
m = 1e-3
proton = charge(1e-6,0+0j,0+0j,-1+0j)
electron = charge(-1e-6,0+0j,0+0j,1+0j)
dt = 0.00001
t=0
last = 'inf'
while True:
    coloumb(proton,electron,dt)
    s = dist(proton.pos,electron.pos)
    if s > last:
        print t
        break
    else:
        last = s
    t+=dt