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Problem with an IVP

  1. Sep 26, 2004 #1
    I'm trying to solve an IVP problem but have not had a lot of Diff Eq to really understand this.

    The IVP is

    dx/dt = -(1-(1/x))*(1/(x^.5)) , x>1
    x(0) = 2

    I guess some of the things I'm looking for with this is:
    Any unique solutions
    equilibrium points
    Initial velocity
    distance increasing or decreasing

    Any additional help on how to solve this would be very much appreciated.

    Thanks in advance.
  2. jcsd
  3. Sep 26, 2004 #2


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    Science Advisor

    We can write dx/dt= -(1-(1/x))*(1/x^.5)) as -(x- 1)/x^1.5)

    There is not t on the right so we can write this in differential form as
    - x^1.5/(x-1)dx= dt.

    Now integrate both sides to find the solutions.

    The rest are even easier. An "equilibrium point" is a point where x doesn't change: the derivative is 0: dx/dt = -(1-(1/x))*(1/(x^.5)) = 0. Solve that equation for x.

    distance (x) is increasing where the derivative is positive and decreasing where it is negative. Since you have a formula for dx/dt, that should be as easy as determining where it was 0.
  4. Sep 26, 2004 #3
    I'm sorry, I guess I don't see how -(1-(1/x))*(1/x^.5)) is the same as -(x- 1)/x^1.5). I've been straining with with algebra but I just don't see that.

    If it is -(x- 1)/x^1.5 then I'm at a loss for how to integrate this.

  5. Sep 27, 2004 #4


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    Subtract the fractions: 1- 1/x= x/x- 1/x= (x-1)/x. Now multiply by (1/x.5): (x-1)/(x*x.5)= (x-1)/(x1.5). All that's missing now is the negative sign.

    -(x-1)/x1.5= -(x/x1.5-1/x1.5)= -x1-1.5+x-1.5= -x-.5- x-1.5. Can you integrate that?
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