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! 

let $\frac{dy}{dx} = w.$ then you have a system of two differential equations $$\frac{dy}{dx} = w, \frac{dw}{dx} = \frac{1}{y^3}$$  from these two, you can get a differential equation $$\frac{dw}{dy} = \frac{1}{y^3}$$ which can be integrated to give you $$w = \frac{dy}{dx} = \frac{1}{2}(C -\frac{1}{y^2})$$ now you separate the variables to get $$\frac{y^2dy}{Cy^2 - 1} = \frac{dx}{2}$$ 
you can use partial fractions to find an integral for $y.$
