1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?