boundedness of solutions of the equation $x''=-\ln(x)-1$

Question

Consider the equation $x''=-\ln(x)-1$. Show that every solution of the equation is a bounded function.

I understand that this is an instance of Newton's equations (?). So, consider the system

$$\begin{cases} x'=y =N(x,y)\\ y'=-\ln(x)-1=-M(x,y)\\ \end{cases} $$ This is an exact equation and the potential function is $$const\equiv E(x,y)=\frac{y^2}{2}+\int\limits_{x_0}^{x}(\ln(t)+1)dt=\frac{y^2}{2}+x\ln(x)$$

How do I deduce boundedness from the above ?

Edit: Can we multiply by $x'$ instead to get (?) $$x''x'+x'\ln(x)+x'=0\implies \frac{(x')^2}{2}+x\ln(x)=c$$ and say that since the component $\frac{(x')^2}{2}$ is non-negative, $x$ cannot grow unboundedly.

Answer 1

Multiplying for $x'$ we have

$$ x' x'' + \log(x)x'+x'=0 $$

and integrating

$$ \frac 12(x')^2+x\log(x) = C_0 $$

then

$$ x' = \pm\sqrt{C_0-x\log(x)} $$

but

$$ \log(x) \le x\log(x) \le x^2 $$

so the DE is separable and comparing we have

$$ \frac{dx}{\sqrt{C_0-{x^2}}}\ge \frac{dx}{\sqrt{C_0-x\log(x)}}\ge \frac{dx}{\sqrt{C_0-\log(x)}} $$

or

$$ \tan ^{-1}\left(\frac{x \sqrt{C_0-x^2}}{x^2-C_0}\right)\le \int\frac{dx}{C_0-x\log(x)}\le \sqrt{\pi } \left(e^{C_0}\right) \text{erf}\left(\sqrt{C_0-\log(x)}\right) $$

Answer 2

Yes, the level curve for $c>0$ is bounded by $$|y|\le\sqrt{2c}$$ and with the Lambert-W function $$x\le e^{W(c)}=c/W(c).$$

It still needs to be shown that no solution reaches $x=0$ in finite time.