Making sense of a 2nd-order nonlinear ODE with *2 unknown functions* and known asymptotic behavior I have the following nonlinear ODE relating the functions $A(r)$ and $B(r)$:
$-\frac{A''}{A} + \frac{A'^2}{2A^2} - \frac{A'B'}{AB} + \frac{B'^2}{2B^2} - \frac{2B''}{B} + \frac{2}{B} = k$
with $k$ a known constant.  I do know one solution:
$A_0(r) = -\cos^2{\frac{r}{\alpha}},\ \ \ B_0(r) = \alpha^2 \sin^2{\frac{r}{\alpha}}$
for a known constant $\alpha$.  In this case $r \in [0,\frac{\pi}{2}\alpha)$ and I believe $k$ works out to be $\frac{12}{\alpha^2}$.
Furthermore, the solution I'd like to find (if there is one) should only differ significantly from $A_0, B_0$ for very small $r$, ie. $r \ll \alpha$.  In other words,
$A(r) = A_0(r) + A_1(r),\ \ \ |A_1(r)| \ll |A_0(r)|\ \text{ unless }\ r \ll \alpha \\
B(r) = B_0(r) + B_1(r),\ \ \ |B_1(r)| \ll |B_0(r)|\ \text{ unless }\ r \ll \alpha$
Also, $A(r) \le 0$ and $B(r) \ge 0$ over the whole domain.  Finally,  $B(0) = 0$ (hence $B_1(0) = 0$) -- and I suspect $B(r)$ may have one more zero, if that helps narrow it down.
With two unknown functions, I guess there's no hope of solving it outright.  But is there a way to turn this into a non-differential equation relating $A$ and $B$?  Or at least a simpler differential equation?  Or to use the $A_0, B_0$ baseline as a guide for a numerical technique or series expansion or something, just to get some better understanding of the solutions that are possible?
 A: This isn't quite an answer, but I thought I'd share with you what I've got since no one else has.
Starting from your ODE, \begin{align}
-\frac{A''}{A}-\frac{1}{2}\left(\frac{A'}{A}-\frac{B'}{B}\right)^2-2\frac{B''}{B}+\frac{2}{B}=k,
\end{align}
I'll assume $B$ is a function of $A$ ($B(A)$), so via the chain rule :
\begin{align}
B'&=B_AA'\\
B''&=B_AA''+B_{AA}A'^2
\end{align} where $B_A$ is short for d$B/$d$A$. This turns the ODE into
\begin{align}
-\left(\frac{1}{A}+\frac{2B_A}{B}\right)A''+\left(\frac{1}{2}\left(\frac{1}{A}-\frac{B_A}{B}\right)^2-\frac{2B_{AA}}{B}\right)A'^2=k-\frac{2}{B}.
\end{align}
Note that we now have an ODE of only one function, $A$, since the $B$'s are unkown functions of $A$. If we divide by the the $A''$ coefficient and rename the functions of $A$ the ODE becomes
\begin{align}
A''+m(A)A'^2=n(A).
\end{align}
This equation has a nice trick to linearize and solve it, let
\begin{align}
A'^2&=C\\
A''&=\frac{1}{2}C_A,
\end{align}
you can verify this using the chain rule. So then the equation becomes
\begin{align}
\frac{1}{2}C_A+m(A)C=n(A),
\end{align}
using an integrating factor, where I've written $\smallint m(A)\mathrm dA=M$ for short,
\begin{align}
\frac{\mathrm d}{\mathrm dA}(e^{2M}C)&=2e^{2M}n(A)\\
C=A'^2&=e^{-2M}\left(c_1+2\int e^{2M}n(A)\mathrm dA\right).
\end{align}
This is separable as well. The resulting integral would give a separable equation,
\begin{align}
\int\frac{e^M \mathrm dA}{\sqrt{c_1+2\smallint e^{2M}n(A)\mathrm dA}}=r+c_2.
\end{align}
The problem lies withing solving these two integrals of unknown functions of $A$, ($B(A)$), though $M$ and $n$ contain derivatives of $B$ which is why I think that maybe you could do it. If the two integrals can be solved, then you'd end up with an equation of $A$, $B$ and $r$, which I believe is what you want.
