The differential equation $r''-w^2r'+\frac{a}{r^2}=0$ I have a problem with finding solution for differential equation:
$$r''-w^2r'+\frac{a}{r^2}=0$$
The term $\frac{a}{r^2}$ is messing simple solution. I'll be very grateful for help.
 A: I'll assume the independent variable is $x$, and $w$ and $a$ are constants. Maple can solve this differential equation by transforming it to an Abel differential equation. But there is no "simple solution". The solution is
$$\int ^{r \left( x \right) }\! \left( {\it RootOf} \left( -2\,{{\rm Bi}
\left(1/4\,{\frac {{2}^{2/3} \left( {t}^{3}{w}^{4}-2\,{\it Z}\,{t}^{
2}{w}^{2}+{{\it Z}}^{2}t-2\,a \right) }{w\sqrt [3]{w{a}^{2}}t}}
\right)}c_{{1}}ta{w}^{2}+2\,{{\rm Bi}\left(1/4\,{\frac {{2}^{2/3}
 \left( {t}^{3}{w}^{4}-2\,{\it Z}\,{t}^{2}{w}^{2}+{{\it Z}}^{2}t-2
\,a \right) }{w\sqrt [3]{w{a}^{2}}t}}\right)}c_{{1}}{\it Z}\,a-2\,{2
}^{2/3} \left( w{a}^{2} \right) ^{2/3}{{\rm Bi}^{(1)}\left(1/4\,{
\frac {{2}^{2/3} \left( {t}^{3}{w}^{4}-2\,{\it Z}\,{t}^{2}{w}^{2}+{{
\it Z}}^{2}t-2\,a \right) }{w\sqrt [3]{w{a}^{2}}t}}\right)}c_{{1}}+2
\,{2}^{2/3} \left( w{a}^{2} \right) ^{2/3}{{\rm Ai}^{(1)}\left(1/4\,{
\frac {{2}^{2/3} \left( {t}^{3}{w}^{4}-2\,{\it Z}\,{t}^{2}{w}^{2}+{{
\it Z}}^{2}t-2\,a \right) }{w\sqrt [3]{w{a}^{2}}t}}\right)}+2\,{
{\rm Ai}\left(1/4\,{\frac {{2}^{2/3} \left( {t}^{3}{w}^{4}-2\,{\it Z
}\,{t}^{2}{w}^{2}+{{\it Z}}^{2}t-2\,a \right) }{w\sqrt [3]{w{a}^{2}}
t}}\right)}ta{w}^{2}-2\,{{\rm Ai}\left(1/4\,{\frac {{2}^{2/3} \left( {
t}^{3}{w}^{4}-2\,{\it Z}\,{t}^{2}{w}^{2}+{{\it Z}}^{2}t-2\,a
 \right) }{w\sqrt [3]{w{a}^{2}}t}}\right)}{\it Z}\,a \right) 
 \right) ^{-1}{dt}-x-c_{{2}}=0
$$
where $Ai$ and $Bi$ are Airy functions and $Ai^{(1)}$ and $Bi^{(1)}$ are their derivatives, and $RootOf(F(Z))$ means a solution of $F(Z) = 0$. 
