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First order ODE.

  1. Nov 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of 2y' + y - (2y')*ln(y') = 0

    2. Relevant equations


    3. The attempt at a solution
    I have no idea how to deal with this i mean none of the first order techniques work and it's mainly because I don't know how to deal with the ln(y').

    I tried seperating the ln to lndy - lndx but it didn't do any good.
     
    Last edited: Nov 25, 2013
  2. jcsd
  3. Nov 25, 2013 #2
    I think a reasonable first step is to differentiate the whole equation once. That should make the logarithm term more approachable.
     
  4. Nov 25, 2013 #3
    Seemed like a good idea but when I differentiated I feel like I complicated it even more by increasing the order

    dy/dx (2y' + y - (2y') * ln(y') = 0)

    2y" + y' - 2y" * ln(y') - 2y'*y''/y' = 0
    = 2y" + y' - 2y" * ln(y') - 2y''
    = y' - 2y" * ln(y')
     
    Last edited: Nov 25, 2013
  5. Nov 25, 2013 #4
    That's how it may seem, but look closer. It's actually a first order ODE for y', and now the logarithm term makes much more sense. There's an obvious substitution you can make to simplify the equation further.
     
  6. Nov 25, 2013 #5
    r = y'
    dr = y"

    r - 2dr * ln(r) = 0

    I feel like something is wrong here maybe I differentiated r incorrectly?
     
  7. Nov 25, 2013 #6

    Mark44

    Staff: Mentor

    In the first line above, your notation indicates that you are multiplying an equation by dy/dx. What you're actually doing is taking the derivative of each side of an equation.

    This notation -- dy/dx(x2) -- means the derivative of some unknown function times x2. Without knowing what dy/dx is, it can't be further simplified.

    This notation -- d/dx(x2) means the derivative, with respect to x, of x2, which is 2x.

    Your 3rd and 4th lines should be their own equations, with 0 on the right side.
     
  8. Nov 25, 2013 #7
    Sorry you're right.

    2y" + y' - 2y" * ln(y') - 2y'*y''/y' = 0
    2y" + y' - 2y" * ln(y') - 2y'' = 0
    y' - 2y" * ln(y') = 0
     
  9. Nov 25, 2013 #8

    Mark44

    Staff: Mentor

    Yes, instead of dr = y'' you should have dr/dy = y'' or r' = y''.

    Using the tip from clamtrox, if you let u = y', so u' = y''. This will give you a first order DE in u that is separable.
     
  10. Nov 26, 2013 #9
    So it would become . . .

    r dy - 2 (ln(r)) dr = 0

    since r = y'

    int ( y'dy) - 2 int (ln(r)) dr) = 0

    Is that right?
     
    Last edited: Nov 26, 2013
  11. Nov 26, 2013 #10
    No, that makes no sense. Let's say y is a function of a variable called x. Then y' = dy/dx. This is also a function of x. Now you say r = y'. Also r is a function of x. So your equation is r - 2 dr/dx * ln(r) = 0. Now you can move all r's on one side, and all x's on the other side, and integrate.
     
  12. Nov 27, 2013 #11
    So after integrating r - 2 dr/dx * ln(r) = 0

    I get x - 2ln(ln(r)) = 0

    x - 2ln(ln(y')) = 0

    x = ln(ln(y')^2)

    e^x = (ln(y'))^2

    I got stuck here not sure how to seperate the y.

    Sorry about replying so late every time it's cause of the time zone.
     
  13. Nov 28, 2013 #12
    Something goes wrong here. The integration is very straightforward if you substitute z = ln(r).
     
  14. Nov 29, 2013 #13
    This was after the integration.

    r - 2 dr/dx * ln(r) = 0

    This is what I integrated.
     
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