<s>[INST] Differential equation $y'' \cdot y^3 = 1$ I use these substitutions $y'=p(y)$ and $y'' = p' \cdot p$ to solve the equation, thus I have the consequence of the solution's steps:
$$ p'py^3 = 1 \implies p'p = \frac{1}{y^3} \implies \frac {dp}{dy} p = \frac {1}{y^3} \implies \int p dp = \int \frac{1}{y^3} dy \implies \\ p = \sqrt{C_1 - \frac {1}{y^2}}$$
Then I try to reverse my substitution and get:
$$ y' = \sqrt{C_1 - \frac {1}{y^2}}$$
What kind of differential equations is it? How do I have to solve it?
 [/INST] Multiply both sides by $2y'/y^3$. Then
$$ 2y'y'' = \frac{y'}{y^3}, $$
and we can recognise both sides as derivatives. Integrating once,
$$ y'^2 = A^2-\frac{1}{2y^2} = \frac{A^2 y^2-1}{y^2}, $$
and then you can rearrange to
$$ \frac{y^2y'^2}{A^2 y^2-1} = 1, $$
and then take a square root and integrate:
$$ x_0 \pm x = \int \frac{y \, dy}{\sqrt{A^2 y^2-1}} = \frac{1}{A^2} \sqrt{A^2y^2-1}, $$
which you can rearrange to find $A^4 (x_0 \pm x )^2 + 1 = A^2 y^2$, (which is a couple of hyperbolae).
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