I'm trying to solve the following differential equation since I'm trying to solve for r(z)=r:
$\frac {d^2r(z)}{dz^2} = \frac {klr^{l-1}} {(n + kr^l)}$
And below is my partial solution:
$\frac {d^2r}{dz^2} = {klr^{l-1}} {(n + kr^l)}^{-1}$
$\frac {d^2r}{dz^2} = \frac{klr^{l-1}}{n} {(1 + \frac{kr^l}{n})}^{-1}$
$\frac {d^2r}{dz^2} = \frac{klr^{l-1}}{n} {\sum_{m}(-1)^m (\frac {kr^l}{n})^m }$
$\frac {d^2r}{dz^2} = \frac{kl}{n} {\sum_{m}(-1)^m (\frac {k}{n})^m (r^{lm+l-1})}$
And this is the part where I'm stuck. The first part of my solution had the goal of trying to move r(z) or remove any r from the denominator so that I could solve the differential equation. But that's all I've been able to do because I'm not sure how to solve it since it has a series that has to be solved with the differential equation. I'm pretty lost with solving this part.
*Side note - This condition cannot be applied: ($\frac {kr}{n} << 1$)
Not a solution
$$\frac {d^2r(z)}{dz^2} = \frac {klr^{l-1}} {(n + kr^l)}$$ $$r''r' = \frac {r'klr^{l-1}} {(n + kr^l)}$$ $$\frac12(r'^2)=\int \frac {r'klr^{l-1}} {(n + kr^l)}dz$$ $$\frac12(r'^2)=\int \frac {klr^{l-1}} {(n + kr^l)}dr$$ $$\frac12(r'^2)=\int \frac {du}{u} \text { ,where } u=(n + kr^l)$$ $$(r'^2)=2\ln|(n + kr^l)|+K$$ $$r'=\pm \sqrt{2\ln|(n + kr^l)|+K}$$ Not that easy to solve indeed...