Analytical approach to this 2nd order ODE? Is there any analytical approach to solving an ODE of the form below? It could be more elegant than numerical approximation...
$$y''-\frac{a}{y^3}+b=0,\quad\textrm{where $a$ and $b$ are constants.}$$
If anyone has any pointers it would be greatly appreciated!
 A: In my opinion, a simple way is to switch variables
$$y''-\frac{a}{y^3}+b=0\quad\implies \quad -\frac{x''}{[x']^3}-\frac{a}{y^3}+b=0$$ Reduction of order $$p=x' \quad\implies \quad \frac{p'}{p^3}+\frac{a}{y^3}-b=0\quad\implies \quad p=x'=\pm \frac{y}{\sqrt{-a+c_1 y^2-2 b y^3}}$$ Now, writing
$$-a+c_1 y^2-2 b y^3=-2b(y-r_1)(y-r_2)(y-r_3)$$  you will face a few elliptic integrals.
The problem would, for sure, be its inverse.
A: Another option is
Solve
\begin{gather*}
\boxed{y^{\prime \prime}-\frac{a}{y^{3}}+b=0}
\end{gather*}
Writing the ode as
\begin{align*}
              y^{\prime \prime}&=\frac{a -y^{3} b}{y^{3}}
\end{align*}
Multiplying both sides by $y^{\prime}$ gives
\begin{align*}
          y^{\prime} y^{\prime \prime}&=\frac{\left(a -y^{3} b \right) y^{\prime}}{y^{3}}
        \end{align*}
Integrating both sides w.r.t. $x$ gives
\begin{align*}
          \int{y^{\prime} y^{\prime \prime}\, \mathrm{d}x}  &=\int{\frac{\left(a -y^{3} b \right) y^{\prime}}{y^{3}}\, \mathrm{d}x}\\ 
          \int{y^{\prime} y^{\prime \prime}\, \mathrm{d}x}  &=\int{\frac{a -y^{3} b}{y^{3}}\, \mathrm{d}y} \tag{1}                
\end{align*}
But
$$
        \int{y^{\prime} y^{\prime \prime}\, \mathrm{d}x} = \frac{1}{2} \left(y^{\prime}\right)^2
$$
And
$$
          \int{\frac{a -y^{3} b}{y^{3}}\, \mathrm{d}y} = -b y-\frac{a}{2 y^{2}}
$$
Therefore  equation (1)  becomes
\begin{align*}
           \frac{1}{2} \left(y^{\prime}\right)^2  &=-b y-\frac{a}{2 y^{2}} + c_2
        \end{align*}
Where $c_2$ is an arbitrary constant of integration.
This is first order ODE separable ode which is now solved for $y$.
Solving for $y^{\prime}$ gives 2 ode's
\begin{align*}
   y^{\prime}&=\frac{\sqrt{-2 y^{3} b +2 c_{2} y^{2}-a}}{y}\tag{1A} \\ 
y^{\prime}&=-\frac{\sqrt{-2 y^{3} b +2 c_{2} y^{2}-a}}{y}\tag{2A} 
\end{align*}
Solving (1A) only (as 2A is similar)
\begin{align*}
           \frac{y}{\sqrt{-2 y^{3} b +2 y^{2} c_{2}-a}}\mathop{\mathrm{d}y}  &= \mathop{\mathrm{d}x}\\   
            \int \frac{y}{\sqrt{-2 y^{3} b +2 y^{2} c_{2}-a}}\mathop{\mathrm{d}y}  &= x +c_{1}
\end{align*}
Using Fricas integrator, the above becomes
\begin{align*}
\frac{\sqrt{2}}{3 \, \sqrt{-b} b} \Delta  &= \left(x +c_{1}\right)
\end{align*}
Where
\begin{equation}
\begin{aligned}
\Delta ={} & c_{2} {\operatorname{weierstrassPInverse}}\left(\frac{4 \, c_{2}^{2}}{3 \, b^{2}}, -\frac{2 \, {\left(27 \, a b^{2} - 4 \, c_{2}^{3}\right)}}{27 \, b^{3}}, \frac{3 \, b y - c_{2}}{3 \, b}\right) - \\
       &\quad  3 \, b {\operatorname{weierstrassZeta}}\left(\frac{4 \, c_{2}^{2}}{3 \, b^{2}}, -\frac{2 \, {\left(27 \, a b^{2} - 4 \, c_{2}^{3}\right)}}{27 \, b^{3}}, {\operatorname{weierstrassPInverse}}\left(\frac{4 \, c_{2}^{2}}{3 \, b^{2}}, -\frac{2 \, {\left(27 \, a b^{2} - 4 \, c_{2}^{3}\right)}}{27 \, b^{3}}, \frac{3 \, b y - c_{2}}{3 \, b}\right)\right)
\end{aligned}
\end{equation}
Same for second ode.
Fricas code
  r:=integrate(y/sqrt(-2*y^3*b+2*y^2*c-a),y);

 (3)
  "((-6)*b*weierstrassZeta((4*c^2)/(3*b^2),(8*c^3+(-54)*a*b^2)/(27*b^3),weierst
  rassPInverse((4*c^2)/(3*b^2),(8*c^3+(-54)*a*b^2)/(27*b^3),(3*b*y+(-1)*c)/(3*b
  )))+2*c*weierstrassPInverse((4*c^2)/(3*b^2),(8*c^3+(-54)*a*b^2)/(27*b^3),(3*b
  *y+(-1)*c)/(3*b)))/(3*b*((-2)*b)^(1/2))"

