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Math Help - Differential Equation

  1. #1
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    Differential Equation

    Given dy/dx = -\frac{8y^2}{(x-1)^3} at point (x,y)
    and the curve goes through (2, -1/3)

    Prove the equation of the curve,

    y=\frac{x^2-2x+1}{x^2-2x-3}
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Separate the variables and integrate...

    \displaystyle \int \dfrac{1}{y^2}\ dy =-8 \int \dfrac{1}{(x-1)^3}\ dx

    Then, use the values of (2, -1/3) to find the value of the constant of integration.

    Or, you can find directly:

    \displaystyle \int^y_{-\frac13} \dfrac{1}{y^2}\ dy =-8 \int^x_2 \dfrac{1}{(x-1)^3}\ dx

    EDIT: You should know by now that you should post what you have tried.
    Last edited by Unknown008; November 17th 2010 at 08:38 AM. Reason: Typo about 'integrate'
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  3. #3
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    Have you tried anything your self? To show that a given function satisifies a differential equation is typically much simpler than actually solving the equation- just differentiate the given y and show that it satisfies the equation.

    However, in this case the equation "separates" as y^{-2}dy= -\frac{8 dx}{(x- 1)^3} and those are easy to integrate.
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  4. #4
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    i have

    \displaystyle \dfrac{-1}{y} =\dfrac{-4}{(x-1)^2}-1
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  5. #5
    MHF Contributor Unknown008's Avatar
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    EDIT: Hm... I'm not getting this. Either I made a mistake or you made one.

    Make y the subject now, then expand the square that you have.
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  6. #6
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    Got it. Ha you guys didnt really help.
    It was just the rearrange i got a bit confused with tbh.
    but thank you as always.
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