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

  1. Jan 27, 2008 #1
    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?
     
  2. jcsd
  3. Jan 27, 2008 #2
    You have

    [tex]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[/tex]

    Let
    [tex]y^2=\frac{z}{2\,C}[/tex]
    and the 1st integral can be easily computed.
     
  4. Jan 27, 2008 #3
    That's where it was, I was missing the part of the derivative for integration. Thanks!
     
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