How to solve a second ODE involving the Lorentzian function, possibly using analytical methods? I'm trying to solve a second ODE involving the Lorentzian function, wherein I'm trying to solve for r(z). To see the form of the Lorentzian function (which I modified slightly for my research) and its derivative, here is a link: http://mathworld.wolfram.com/LorentzianFunction.html
From the above link, I changed L to n and let x = r and xo=m. 
$$n(r) =  \frac {1}{\pi} \frac {\frac{1}{2} \Gamma}{(r-m)^2+({\frac{1}{2} \Gamma})^2}$$                    
$$\frac {dn}{dr}=n'(r)=-\frac {16(r-m) \Gamma}{\pi [4(r-m)^2 + (\Gamma)^2]^2} $$
The second order ODE that uses the above modified versions of the Lorentzian function is shown below:
$$\frac {d^2r}{dz^2} = \frac {1}{n(r)} \frac {dn}{dr}$$
I inputted the functions into the above equation and simplified it, obtaining:
$$\frac {d^2r}{dz^2} = - \frac {2[(r-m)^3 + (r-m)(\frac {1}{4} \Gamma^2)]}{[(r-m)^4+8(r-m)^2(\Gamma^2)+ \Gamma^4]}$$
And this is where I had to stop because I couldn't solve it any further by hand. Based on what I've tried, it cannot be solved using the usual second ODE solutions. I'm trying out at analytic methods but I'm still having difficulty with this and am still self-studying analytic methods for this. 
Edit: I tried approximations using numerical methods but it didn't work out but feel free to still try that method. 
 A: While I'm not sure what is $z$, I will show how to solve the provided equation:
$$\frac {d^2r}{dz^2} = - \frac {2[(r-m)^3 + (r-m)(\frac {1}{4} \Gamma^2)]}{[(r-m)^4+8(r-m)^2(\Gamma^2)+ \Gamma^4]}$$
First, we introduce a new function:
$$f(z)=\frac{r(z)-m}{\Gamma}$$
We can rewrite the equation as:
$$\frac {d^2 f}{dz^2} = - \frac {2f^3 + \frac {1}{2}f}{f^4+8f^2+ 1}$$
We can consider the inverse function $z(f)$, then we have:
$$\frac {d^2 f}{dz^2} =- \frac{z''(f)}{(z'(f))^3}$$
So we can write:
$$\frac {z''}{z'^3} = \frac {2f^3 + \frac {1}{2}f}{f^4+8f^2+ 1}$$
Now introducing another function:
$$z'(f)=y(f)$$
We have a 1st order separable ODE:
$$\frac {y'}{y^3} =  \frac {2f^3 + \frac {1}{2}f}{f^4+8f^2+ 1}$$
Integrating we have:
$$-\frac{1}{2 y^2}=\int \frac {2f^2 + \frac {1}{2}}{f^4+8f^2+ 1} fdf$$
Setting $f^2=p$, the integral becomes:
$$\int \frac {\left(p + \frac {1}{4} \right) dp}{p^2+8p+ 1}=\int \frac {\left(p + \frac {1}{4} \right) dp}{(p+4)^2-15}=\int \frac {(p+4)dp}{(p+4)^2-15} -\frac{15}{4} \int \frac {dp}{(p+4)^2-15} =$$
Setting $p+4= \sqrt{15} q$ we have:
$$=\int \frac {q dq}{q^2-1} -\frac{\sqrt{15}}{4} \int \frac {dq}{q^2-1}=\frac{1}{2} \ln |1-q^2|-\frac{\sqrt{15}}{8} \ln \left| \frac{1-q}{1+q} \right|+const$$
Getting back to our ODE we have:
$$\frac{1}{y^2}=\frac{\sqrt{15}}{4} \ln \left| \frac{1-q}{1+q} \right|-\ln |1-q^2|+const$$
Where:
$$q=\frac{f^2+4}{\sqrt{15}}$$
Now we can explicitly write $y(f)$, then remember that $y=z'$, integrate the resulting separable 1st order ODE and get implicitly our function $f(z)$ or equivalently $r(z)$.
The result is going to be ugly, so I won't continue further.
