What is the solution to this non-linear second order differential equation? I'm trying to solve the following non-linear second order differential equation:
$$\tag{1}
\frac{d\, }{dx} \Bigl( \frac{1}{y^2} \, \frac{dy}{dx} \Bigr) = -\, \frac{2}{y^3},
$$
where $y(x)$ is an unknown function on the real axis.  I already know the "trivial" solution $y(x) = y_0 \pm x$.  The solution I'm looking may involve trigonometric functions, but I'm not sure.  Take note that we may pose
$$\tag{2}
u = \frac{1}{y},
$$
so that (1) takes another form:
$$\tag{3}
\frac{d^2 u}{dx^2} - 2 \, u^3 = 0.
$$
So what is the non-linear solution $y(x)$?  Or $u(x)$?

Solve
\begin{align*}
              u^{\prime \prime}&=2 u^{3}
\end{align*}
Multiplying both sides by $u^{\prime}$ gives
\begin{align*}
          u^{\prime} u^{\prime \prime}&=2 u^{3} u^{\prime}
        \end{align*}
Integrating both sides w.r.t. $x$ gives
\begin{align*}
\int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x}  &=\int{2 u^{3} u^{\prime}\, \mathrm{d}x}\\ 
\int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x}  &=\int{2 u^{3}\, \mathrm{d}u} \tag{1}                
\end{align*}
But
$$
        \int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x} = \frac{1}{2} \left(u^{\prime}\right)^2
$$
Hence equation (1)  becomes
\begin{align*}
        \frac{1}{2} \left(u^{\prime}\right)^2  &=\int{2 u^{3}\, \mathrm{d}u} \tag{2}                
        \end{align*}
But
$$
          \int{2 u^{3}\, \mathrm{d}u} = \frac{u^{4}}{2}
$$
Therefore  equation (2)  becomes
\begin{align*}
           \frac{1}{2} \left(u^{\prime}\right)^2  &=\frac{u^{4}}{2} + c_2
        \end{align*}
Where $c_2$ is an arbitrary constant of integration.
This is first order ODE which is now solved for $u$.
Solving for $u^{\prime}$ gives
\begin{align*}
   u^{\prime}&=\sqrt{u^{4}+2 c_{2}}\tag{1} \\ 
u^{\prime}&=-\sqrt{u^{4}+2 c_{2}}\tag{2} 
\end{align*}
Let just solve (1) as (2) is similar.
\begin{align*}
           \frac{1}{\sqrt{u^{4}+2 c_{2}}}\mathop{\mathrm{d}u}  &= \mathop{\mathrm{d}x}\\   
            \int \frac{1}{\sqrt{u^{4}+2 c_{2}}}\mathop{\mathrm{d}u}  &= \int \mathop{\mathrm{d}x}\\ 
            \int \frac{1}{\sqrt{u^{4}+c_{3}}}\mathop{\mathrm{d}u}  &= x +c_{1}
\end{align*}
Using the computer, integrating the left side above gives the solution
\begin{align*}
\frac{\sqrt{1-\frac{i u^{2}}{\sqrt{c_{3}}}}\, \sqrt{1+\frac{i u^{2}}{\sqrt{c_{3}}}}\, \operatorname{EllipticF}\left(u \sqrt{\frac{i}{\sqrt{c_{3}}}}, i\right)}{\sqrt{\frac{i}{\sqrt{c_{3}}}}\, \sqrt{u^{4}+c_{3}}} =  x +c_{1}
\end{align*}
Similar solution for the second ode.
