Reducible Second Order Differential Equation: Ind. and First Derivative Missing

[JTemple]

Homework Statement

Solve the reducible 2ODE. Assume x, y and/or y' positive where helpful.

y^3 * y'' = 1

The Attempt at a Solution

Well, I tried what I normally would do for x being missing.

p = (dy/dx); y'' = p'p = (dp/dy)(dy/dx)

So

y^3 p'p = 1

p(dp/dy) = y^(-3)

Separate, Integrate

\intpdp= \inty^-3dy

(1/2)p^2 = -(1/2)y^-2 + C

p = (2C - (y^-2))^(1/2)

1/p = dx/dy, so

dx = (2C - (y^-2))^(1/2) dy

And then I would try and integrate that. I can't get the right side (I thought arcsin, but aren't I missing the du (Calc II memory bad), or do I add 0 in the form of + something - something in the numerator? Or is my overall approach off?

 

[Rainbow Child]

You have

p=\sqrt{2\,C-\frac{1}{y^2}}\Rightarrow \frac{d\,y}{d\,x}=\frac{\sqrt{2\,C\,y^2-1}}{y}\Rightarrow \int\frac{y\,d\,y}{\sqrt{2\,C\,y^2-1}}=\int d\,x

Let
y^2=\frac{z}{2\,C}
and the 1st integral can be easily computed.

 

[JTemple]

That's where it was, I was missing the part of the derivative for integration. Thanks!