I am considering the following non-linear 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 $T< 1$ a fixed arbitrarily small number. I would like to show that there exists an initial condition $\dot y(T) = c$ such that $$0 \le y(x) \le C\frac{x}{|\ln(x)|^2}.$$ Is there any tool I could use to proof that kind of estimates ?
Actually, by assuming that $y(x) = x/|\ln(x)|^2$, then we get \begin{align} \log(x) - 2\log y(x) &= |\log(x)| + 4 \log|\log(x)|\\\\ \frac{(\dot y(x))^2}{y(x)} &= \frac{|\ln(x)|^2}{x}\left(\frac{1}{|\ln(x)|^2} + 2 \frac{1}{|\ln(x)|^3} \right)\dot y(x) \\\\ &= \frac{\dot y(x)}{x} + 2\frac{\dot y(x)}{x|\ln(x)|} \end{align}
Therefore, if we consider $T \ll 1$, the terms $|\ln(x)|$ dominates in the first equality and $\frac{\dot y(x)}{x}$ dominates in the second one. Hence, at main order the equation can be approximated by \begin{cases} \ddot y(x)|\ln(x)| - 2\frac{\dot y(x)}{x} = 0 &\text{in }[0,T]\\\\ y(0) = 0\\\\ \dot y(T) = c \end{cases} which is explicitly solved by $$y(x) = \int_0^x \frac{c}{|\ln(s)|^2}ds \quad \Rightarrow \quad y(x) = O\left(\frac{x}{|\ln(x)|^2}\right).$$ I feel that this suggests that it's possible to find and appropriate solution with the desired estimate, but I haven't been able to find a rigorous proof. I cannot think of any tool I could apply in this context that would help me with this. Any idea ?