I'm interested in knowing the asymptotics of solutions to the nonlinear ordinary differential equation \begin{equation*} \begin{array}{ll} y''=1/y^m\tag{*}\\ y(0)=a>0, \text{ and }y'(0)=0. \end{array} \end{equation*}
When $m=3$ $(*)$ has a closed form solution $y=\frac{1}{a}\sqrt{a^4+x^2}$, and $y$ is asymptotic to $\frac{1}{a}|x|$. I have been trying to deduce for which $m$ $(*)$ has a solution which is asymptotically linear (as for $m=3$), and also the constants $c,b$ depending on $m,a$ such that $y$ is asymptotic to $c|x|+b$.
So far I've begun by turning the second order ODE into a first order equation. This follows from letting $z=\frac{dy}{dx}$ and noticing that $(*)$ is equivalent to $z \frac{dz}{dy}=y'' = 1/y^m$, and solving for $z=\frac{dy}{dx}$ to get \begin{equation} (y')^2=\frac{2}{1-m} y^{1-m}+C. \end{equation}
Then we can evaluate $C$ using the initial condition and get for $x\geq 0$ and $y\geq a$ \begin{equation} y' = \left( \frac{2}{m-1} \left(\frac{1}{a^{m-1}} - \frac{1}{y^{m-1}}\right)\right)^{1/2}. \end{equation}
$y$ is a convex function because $y''=1/y^m \geq 1/a^m$, so in particular it is unbounded, so letting $y\to \infty$ in the above expression gives \begin{equation} \lim_{x\to \infty} y'(x) = \left(\frac{2}{(m-1)a^{m-1}}\right)^{1/2} :=c. \end{equation}
This step required that $m>1$, so that's the first constraint on $m$. If $y$ is asymptotic to anything, it will be of the form $y=c|x|+b$ for some constant $b$. Since $0\leq y' \leq c$ it follows that $y$ is bounded above by $c|x|+a$, but it's not clear to me that $y$ is necessarily bounded below by a linear function $c|x|+b$ for some $b$. I've tried get asymptotic bounds on the integral
\begin{equation} \int_a^y \left( \frac{2}{m-1} \left(\frac{1}{a^{m-1}} - \frac{1}{t^{m-1}}\right)\right)^{1/2} \,dt \end{equation}
as $y\to \infty$, but I haven't had much success. If anyone has any suggestions for how to prove the solution $y$ is bounded below by $c|x|+b$ for some $b$ I'd be very grateful.
Update: I haven't proved this yet, but I've made progress by showing that $y$ cannot be asymptotic to $cx-ln(x)$. This would mean that there exists $\alpha>\beta>0$ such that as $x\to \infty$ \begin{equation} cx-\alpha\, ln(x) \leq y(x) \leq cx-\beta \,ln(x). \end{equation} Because $f$ is convex this implies a bound on the derivative. \begin{equation} c-\frac{\alpha}{x} \leq y'(x) \leq c-\frac{\beta}{x}. \end{equation} And then you can show from $(*)$ that $y'''<0$ so $y'$ is concave and this implies a bound on the second derivative. \begin{equation} \frac{\alpha}{x^2} \leq y''(x) \leq \frac{\beta}{x^2}. \end{equation}
Then if $m>2$ we get the following inequality for large enough $x$:
\begin{equation} \frac{1}{y^m} \leq \frac{1}{(cx-ln(x))^m} < \frac{\alpha}{x^2} \leq y''. \end{equation}
This provides us with a contradiction so if $m>2$ then $y$ cannot be asymptotic to $cx-ln(x)$. But this is only one example, and doesn't stop $y$ from being some other sublinear function. One could probably prove a similar thing for $y$ asymptotic to $cx-ln(ln(x))$, but this method will never completely prove that $y$ is asymptotically linear.