Solution to an annoying ODE I'm very sorry if this is a duplicate, but searching for a specific equation is rather difficult. I have encountered the following ODE during my physics research (have a PDE which has a similarity solution), and am wondering whether there's any chance of obtaining an analytic solution to it:
\begin{equation} x y' = -2(A(y) y')' ,\end{equation} for $y(x)$ with $y'=\frac{\mathrm{d}y}{\mathrm{d}x}$, just to be clear.
$A(y) = 1-\zeta y(4-3y)$, for $\zeta \ge 0$, although I suspect that if we find ourselves invoking that we've probably already failed, so maybe treat $A(y)$ as being generic for now. I've fiddled around with it, but without much success. In terms of boundary conditions, well, they're up for grabs, was hoping the solution would tell me a bit more about them; physically $0 \le y \le 1$, and would probably like to prescribe Dirichlet conditions on either side of a finite domain. If it helps, I got to this equation using this: https://arxiv.org/pdf/0710.4000.pdf , around page 5.
Of course, asymptotic solutions as $x\rightarrow 0$ or $x\rightarrow \pm \infty$ would also be very cool, as would advice about how to tackle it numerically if analytic solution attempts prove fruitless.
 A: I found a couple additional facts, so I'll also include what I mentioned in the comments below:
Case 1: $y=const.$ is a solution for any constant, so we omit this solution in what follows.
(Correction to my earlier comment: $y(x)=ax+b$ is not a solution if $a,b\neq 0$)
Case 2: if $\zeta=0$, then $y''(x)=-x y'(x)/2$, and $y(x)=a+b\text{ erf}(x/2)$, where $\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{-\infty}^xe^{-t^2}dt$, and $a,b=const.$
Case 3:  Suppose now $\zeta\neq 0$.  Put $\mu=-4+\frac{3}{\zeta}\in(-4,\infty)$, and change variables by setting
$$
y(x)=\frac{2+z\left(\sqrt{\frac{3}{2\zeta}}x\right)}{3}.
$$
Then $z$ solves
$$
\frac{d}{dt}[(z^2+\mu)z'(t)]+tz'(t)=0.
$$
Case 4: $\mu=0$ ($\zeta=3/4$).  In this case, we have the ODE
$$
\frac{d}{dt}[z^2z'(t)]+tz'(t)=0,
$$
which is dilatation invariant under $t\mapsto at$, $z\mapsto az$.  Consequently, if we set $z(t)=t\,u(\ln|t|)$ and $u'(s)=w(u(s))$, we obtain the following first order ODE for $w(u)$:
$$
w'(u)=-\frac{1}{2}\left(\frac{w^2+u^2}{uw}+\frac{u+w}{u^2w}+\frac{5}{2}\right),
$$
which Mathematica could not solve, unfortunately. 
Case 5: $\mu>0$ ($0<\zeta<3/4$). Putting $z(t)=\mu^{1/2}w(t)$ gives
$$
\frac{d}{dt}[(w^2+1)w'(t)]+tw'(t)=0.
$$
Case 6: $\mu<0$ ($\zeta>3/4$). Putting $z=(-\mu)^{1/2}w$ gives
$$
\frac{d}{dt}[(w^2-1)w'(t)]+tw'(t)=0.
$$
Unlike Case 4, the equations of Cases 5 and 6 have no obvious symmetries, so analytic integration seems impossible for them.  However, we can try asymptotic analysis.  I will only do analysis in $w$, since the analysis in $x$ is involved.
Case A: $|w|\ll 1$.  This will apply for both Cases 5 and 6.  Setting $w=\epsilon f_0+\epsilon^2 f_1+\cdots$ and letting $\epsilon\to 0$, we want $f_0$ to solve
$$
\pm f_0''(t)+tf_0'(t)=0,
$$
where $+$ is Case 5, and - is Case 6.  The solution is $f_0(t)=a+b \text{ erf}(t/\sqrt{2})$ if $+$, and $f_0(t)=a+b \text{ erfi}(t/\sqrt{2})$ if $-$.  Note that the latter blows up at infinity.  In any case,
$$
w(t)=\epsilon f_0(t)+O(\epsilon^2)
$$
is a solution as $\epsilon\to 0$, where $\epsilon$ is an arbitrary small parameter of the same order of magnitude as the initial condition $w(0)$.
Note that this is valid for all $\zeta$ considered, since $\epsilon$ does not depend on $\zeta$.
Case B: $|w|\gg 1$.  This recovers Case 4.  So if you can solve this $\zeta=3/4$ equation, say, numerically, then you obtain solutions for all other values of $\zeta$ as well, provided they satisfy $|w|\gg 1$.
Case C: $|w|\approx 1$, and $\mu<0$ (i.e. Case 6).  The idea is to make the coefficient $w^2-1$ vanish.  Put $w(t)=1+\epsilon f_0(\epsilon^{-1/2}t)+\epsilon^2 f_1(\epsilon^{-1/2}t)+\cdots$ and let $\epsilon\to 0$:
$$
\frac{d}{ds}[2f_0(s)f_0'(s)]+sf_0'(s)=0.
$$
This is invariant under the dilatation $s\mapsto as$, $f_0\mapsto a^2f_0(s)$, so we can put $f_0(s)=s^2 g(\ln|s|)$, $g'(r)=h(g(r))$ as before and obtain a first order ODE for $h=h(g)$.  (Mathematica couldn't solve this one either).
Anyways, I hope some of this helps.  I can't promise the above is typo-free, so I would verify Case A for yourself, since this one might be of some use.
