- #1
JTemple
- 9
- 0
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
[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?