Recent content by j_reez

  1. J

    2nd order nonlinear DE need help bad

    Thank You thanks guys! its very clear to me now :) my book had one very limited example which made it sound like i only need to make one substitution. however, i clearly see it now! thanks, Justin
  2. J

    2nd order nonlinear DE need help bad

    i dont see how that would work....if u=xy' then du/dx=y' +xy'' how would thos substitutions work?
  3. J

    2nd order nonlinear DE need help bad

    this problems driving me crazy :surprised
  4. J

    2nd order nonlinear DE need help bad

    this is the problem: xy'' -x(y')^2 = y' my book says that i need to substitute u=y' and du/dx=y''... so i get: x(du/dx)-xu^2 = u so next the book says i need to separate the x's/dx' to one side and u's/du's to the other. however, i cannot do it am i using the correct technique...
  5. J

    Need help with nonlinear 2nd order DE

    im really not seeing how this can be separated :grumpy:
  6. J

    Need help with nonlinear 2nd order DE

    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?
  7. J

    Need help with nonlinear 2nd order DE

    ok hows this look: x(du/dx) -u^2 = u x(du/dx) = u + u^2 (1/x)dx = (1/u+u^2)du ?
  8. J

    Need help with nonlinear 2nd order DE

    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...
  9. J

    Need help with nonlinear 2nd order DE

    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
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