<s>[INST] 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 $z,x$
 [/INST] HINT
\begin{align*}
y'' + y = \frac{1}{y^{3}} & \Longleftrightarrow y''y' + yy' = \frac{y'}{y^{3}}\\\\
& \Longleftrightarrow (y')^{2} + y^{2} = -\frac{1}{y^{2}} + c\\\\
& \Longleftrightarrow (y')^{2} = \frac{cy^{2} - y^{4} - 1}{y^{2}}\\\\
& \Longleftrightarrow y' = \pm\sqrt{\frac{cy^{2} - y^{4} - 1}{y^{2}}} 
\end{align*}
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