(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

y^3 * y'' = 1

3. 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

[tex]\int[/tex]pdp= [tex]\int[/tex]y^-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?

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

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