<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] Another Way to look at it
$$\frac{d^2y}{d^2x} = \frac{1}{y^3} - y  =>  \frac{d^2y}{d^2x} = \frac{1-y^4}{y^3}   =>$$
$$\frac{y^3}{1-y^4} d^2y = d^2x $$
Now this can be double integrated using the integral of a division formula
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