Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$ Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable
Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equation completely involving $z,x$

Another possibility
Switch variables to make
$$-\frac {x''}{[x']^3}+y=\frac 1 {y^3}$$ Reduction of order $p=x'$ gives a separable equation
$$-\frac {p'}{p^3}=\frac 1 {y^3}-y\implies p=\pm\frac{y}{\sqrt{c_1  y^2-y^4-1}}$$ Now
$$x+c_2=\pm \int \frac{y}{\sqrt{c_1  y^2-y^4-1}}\,dy=\pm \frac 12 \int \frac{dt}{\sqrt{c_1  t-t^2-1}}$$ and you will arrive to some arctangent.
