Behaviour of the solution of a second order ODE

I am currently studying the following second order ODE $$\begin{cases} \ddot y(x)\left(\ln(x) - 2\ln(y(x))\right) - 2\frac{(\dot y(x))^2}{y(x)} = 0 &\text{in }[0,T]\\ y(0) = 0\\ \dot y(T) = c \end{cases}$$ for $$c$$ a positive constant and $$T > 0$$ and arbitrarily small constant. Clearly, solving this ODE explicitly is not doable. But actually for my purpose I don't need a whole solution but rather to understand the behaviour of $$\dot y(x)$$ near $$0$$. I would like to know if $$\lim_{x \to 0} \dot y(x) = 0$$ or not.

Being not really familiar with the tools used in ODE theory, I was wondering if one of you guide me on this topic. I solved the ODE numerically using Maple and the software gave me the following plots

where I fixed $$T = 1/2$$ and the initial conditions set at $$c = 1, 1/2$$ and $$1/3$$ corresponding to $$1, 2$$ and $$3$$ respectively. It seems that the slope at the origin is always strictly positive, whatever initial condition I set (if I decrease $$c$$ or $$T$$, the behaviour remains the same). However, I have absolutely no idea how to do that. Any help?

I initially asked this question Stackexchange Mathematics but as I haven't got any answer I crossposted it.

• I cannot parse "for $c$ a positive constant and $T > 0$ and arbitrarily small constant". Commented Jan 26 at 12:32
• I don't think it's clear that this even has a solution. Commented Jan 26 at 18:55
• I'm skeptical that "Clearly" this cannot be solved explicitly. Commented Jan 26 at 22:56

1. A first quick answer to the question as it is (to know if $$\lim_{x\to0}\dot y(x)=0$$ or not), is no in general, as it is shown by the solution $$y(x)=e\sqrt x$$ (corresponding to $$c=\frac1{\sqrt2}$$ for $$T=\frac12$$).
2. Deciding if for given $$T, c$$ a solution with $$y(0)=0$$ and $$\dot y(T)=c$$ satisfies the condition or not is of course a more difficult task (it is not even clear the issue of existence and uniqueness for these solutions, nor if the information $$\dot y(T)=c$$ is sufficient to decide).
3. A conjugation by the exponential function , that is the substitution $$u(t):=\log(y(e^t))$$, produces an algebraic ODE $$(\ddot u-\dot u)(2u-t)+ (2u-t+2) \dot u^2 =0$$ of whom you want to study the behaviour for $$t\to-\infty$$ (one has $$y(e^t)=e^{u(t)}$$ and $$\dot y(e^t)= e^{u(t)-t} \dot u(t)$$) .
4. The above transformed ODE seems to have several polynomial solutions, that can be found solving the polynomial system on the coefficients that one obtains by imposing the Ansatz to the ODE. These may be useful examples to inspect your problem. There is a linear one, $$u(t):=1+t/2$$ (corresponding to the solution above) a few of degree $$2$$, $$u(t):=a+bt+ct^2$$ where $$a$$ is a real root of $$8a^6-36a^4+24a^3-42a^2-12a-3=0$$ (according to Maple they are $$a=-2.5335..$$ and $$a=-0.1419..$$), and $$b$$ and $$c$$ are certain algebraic expression of $$a$$. It is not clear to me if there is a more efficient way to find these polynomial solutions.
• Nice trick, but I think there is a typo: If I'm not wrong the numerator reads $-2u+t-2$. Commented Jan 26 at 15:08