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Need help with nonlinear 2nd order DE

  1. Nov 8, 2004 #1
    problem: xy'' -x(y')^2 = y'

    what i have so far:

    u=y' and du/dx=y''

    du/dx - u^2 = (1/x)u

    int[(1/u)-u]du = int[1/x]dx

    ln u - (1/2)u^2 = ln x +c

    ok, now is what ive done so far correct? what do i do next?

    ps: i'd like to say hi to everyon :) im new here
  2. jcsd
  3. Nov 9, 2004 #2
    welcome to the board!

    You have made an error here -->
    du/dx - u^2 = (1/x)u
    int[(1/u)-u]du = int[1/x]dx

    What u will have is ...
    du/dx - u^2 = (1/x)u
    du - u^2dx =(u/x)dx

    Can u see the error u made?
    Can u correct it?

    -- AI
  4. Nov 9, 2004 #3
    i must be making a trivial algebraic mistake....as far as i know im supposed to be isolating dx's and x's on one side with u's and du's on the other...which is why i divided through by x. is this not allowed?

    oh boy, i see it...i cant get dx to the other side like that...let me see what i can do
  5. Nov 9, 2004 #4
    ok hows this look:

    x(du/dx) -u^2 = u

    x(du/dx) = u + u^2

    (1/x)dx = (1/u+u^2)du

  6. Nov 9, 2004 #5
    du/dx - u^2 = (1/x)u
    Multiplying throughout by x gives,
    x(du/dx) - u^2*x = u

    -- AI
  7. Nov 9, 2004 #6
    yes that was the form it was in. ive got it down to this:

    [int]dx/x = [int]du/(u(u+1))

    how do i integrate the RHS?
  8. Nov 9, 2004 #7
    think again ...

    ur original equation was,
    xy'' -x(y')^2 = y'
    placing u = y' u get,
    xu'--xu^2 = u

    The way u have separated wont work ...
    Think harder!! :)

    -- AI
    P.S as an aside,
    to integrate 1/(u(u+1))
    u should use partial fractions and integrate
    ofcourse for now it wont apply to this problem
    i am just telling this as it might be helpful somewhere else
  9. Nov 9, 2004 #8
    im really not seeing how this can be separated :grumpy:
  10. Nov 9, 2004 #9
    The thing is that it cannot be separated atleast in this form....

    Substitute u = vx ...
    where v is some function of x ...

    -- AI
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