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 $u(x)$?
 A: 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.
A: Multiplying by $u'$ and regrouping gives an equality of derivatives
$$\frac{1}{2} ((u')^2)' = \frac{1}{2} (u^4)',$$
and integrating gives
$$(u')^2 = u^4 + c .$$
This equation is separable, and rearranging and integrating gives
$$\pm \int \frac{du}{\sqrt{u^4 + c}} = x + d .$$
If $c < 0$, we can write $c = -a^4$ for some $a > 0$, and in the variable $v = \frac{u}{a i}$ the left-hand side is
$$\pm \frac{i}{a} \int \frac{dv}{\sqrt{v^4 - 1}} = \pm \frac{i}{a} F(iv, i) = \pm \frac{i}{a} F\left(\frac{u}{a}, i\right) ,$$
where $F$ is the incomplete elliptic integral of the first kind and where we've absorbed the constant of integration into $d$. Then, we can write $u$ in terms of the Jacobi elliptic function $\operatorname{sn}$:
$$u(x) = \mp a i \operatorname{sn} (a(x + d), i) .$$ The reciprocal of $\operatorname{sn}$ is denoted $\operatorname{ns}$, and so we may write
$$y(x) = \pm \frac{i}{a} \operatorname{ns} (a(x + d), i) .$$
The case $c > 0$ can be handled similarly, and the case $c = 0$ yields the linear solutions $y(x) = y_0 \pm x$.
