Find general solution $y''y^3=1$ I'm having this question in my homework assignment in Linear Algebra and diffrential equation class, and trying to find the general solution for this second ODE.
$$y''y^3 = 1$$
Using substitution I said $p = y'$ and $p' = y''  \rightarrow  \frac{dp}{dx}= \frac{dp}{dy} \times \frac{dy}{dx}$
Then we get $y^3 \frac{dp}{dx}$ now what's the next step should I integrate or differentiate the equation? What's the trick on these type of ODEs?
Thanks in advance! 

Divide by y^3 then multiply by y' and integrate. That gets you to first order. I think the next integration is possible but a bit messy. 
