Can't solve complicated second order differential equation (Poisson-Boltzmann equation) Is there anyone able to solve this second order differential equation? It is the Poisson-Boltzmann equation (found in the field of electrostatics) solved on cylindrical coordinates just on the radial direction.
$$
(\varphi'+r\cdot\varphi'')=A\cdot e^{−(B\cdotφ+C)}
$$
where $\varphi$ is the variable I want to solve the problem for, the derivatives of $\varphi$ are with respect to $r$, while $A,B,C$ are constants.
I have been around it for a long time and can't solve it. What I had tried was to do a variable change by setting a new variable called z for instance which is equal to the exponential term, but I arrive nowhere. 
All the help is much appreciated :)
 A: Let $r=e^s$ ,
Then $s=\ln r$
$\dfrac{d\varphi}{dr}=\dfrac{d\varphi}{ds}\dfrac{ds}{dr}=\dfrac{1}{r}\dfrac{d\varphi}{ds}=e^{-s}\dfrac{d\varphi}{ds}$
$\dfrac{d^2\varphi}{dr^2}=\dfrac{d}{dr}\left(e^{-s}\dfrac{d\varphi}{ds}\right)=\dfrac{d}{ds}\left(e^{-s}\dfrac{d\varphi}{ds}\right)\dfrac{ds}{dr}=\left(e^{-s}\dfrac{d^2\varphi}{ds^2}-e^{-s}\dfrac{d\varphi}{ds}\right)e^{-s}=e^{-2s}\dfrac{d^2\varphi}{ds^2}-e^{-2s}\dfrac{d\varphi}{ds}$
$\therefore e^{-s}\dfrac{d\varphi}{ds}+e^{-s}\dfrac{d^2\varphi}{ds^2}-e^{-s}\dfrac{d\varphi}{ds}=Ae^{−(B\varphi+C)}$
$e^{-s}\dfrac{d^2\varphi}{ds^2}=Ae^{−(B\varphi+C)}$
$\dfrac{d^2\varphi}{ds^2}=Ae^{s−B\varphi-C}$
Let $u=s−B\varphi-C$ ,
Then $\varphi=\dfrac{s−u-C}{B}$
$\dfrac{d\varphi}{ds}=\dfrac{1}{B}-\dfrac{1}{B}\dfrac{du}{ds}$
$\dfrac{d^2\varphi}{ds^2}=-\dfrac{1}{B}\dfrac{d^2u}{ds^2}$
$\therefore-\dfrac{1}{B}\dfrac{d^2u}{ds^2}=Ae^u$
$\dfrac{d^2u}{ds^2}=-ABe^u$
$u=\ln\dfrac{2c_1\text{sech}^2\sqrt{c_1(s+c_2)^2}}{AB}$
$s−B\varphi-C=\ln\dfrac{2c_1\text{sech}^2\sqrt{c_1(s+c_2)^2}}{AB}$
$\varphi=\dfrac{s-C}{B}−\dfrac{1}{B}\ln\dfrac{2c_1\text{sech}^2\sqrt{c_1(s+c_2)^2}}{AB}$
$\varphi=\dfrac{\ln r-C}{B}−\dfrac{1}{B}\ln\dfrac{2c_1\text{sech}^2\sqrt{c_1(\ln r+c_2)^2}}{AB}$
