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A little help with calculus terms.

  1. Mar 11, 2005 #1
    The rate of change of y with respect to x is inversely proportional to the square root of y.
    a)Write a differential equation for the given statement
    b)Solve the differential equation in part a.

    I don't know, but what I've done so far is:
    [tex]({dy/dx}) k=y^{1/2}[/tex]
     
    Last edited: Mar 11, 2005
  2. jcsd
  3. Mar 11, 2005 #2
    I'm not sure, but I think this is it:

    a) (dy/dx)x=1/sqrt(y)

    b)
    (dy/dx)1/x²=y
    -2x/x^4=y
     
  4. Mar 11, 2005 #3

    dextercioby

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    Incorrect.It's INVERSE PROPORTIONALITY.We usually let the constant in the other side of the equality.

    dy(x)/dx~y^{-\frac{1}{2}}=>[tex] \frac{dy(x)}{dx}=ky^{-\frac{1}{2}} [/tex]

    Daniel.
     
    Last edited: Mar 11, 2005
  5. Mar 11, 2005 #4

    dextercioby

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    Sorry,there's no "x" explicitely.Just "y" to a power & its first derivative of "y".


    Daniel.
     
  6. Mar 11, 2005 #5
    why is the x beside the (dy/dx)?
     
  7. Mar 11, 2005 #6
    Nevermind, I think I misunderstood the whole point of the question :/
     
  8. Mar 11, 2005 #7
    ok, but why is the "x" in the left side of the equation, why is there an x at all?
     
  9. Mar 11, 2005 #8

    dextercioby

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    It isn't.It shouldn't be.It was an error from the poster.

    Daniel.
     
  10. Mar 11, 2005 #9
    so the solution should just be [tex] \frac{dy}{dx}=ky^{-\frac{1}{2}} [/tex] ?
     
  11. Mar 11, 2005 #10

    dextercioby

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    Yes.

    Daniel.
     
  12. Mar 11, 2005 #11
    that would mean that y=?
     
  13. Mar 11, 2005 #12

    dextercioby

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    You can find easier "x" as a function of "y".

    Daniel.
     
  14. Mar 12, 2005 #13

    HallsofIvy

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    [tex]\frac{dy}{dx}= ky^{\frac{-1}{2}}[/tex]
    is a "separable equation". Write it as
    [tex]y^{\frac{1}{2}}dy= kdx[/tex]
    and integrate.
     
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