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Thread: Integration... can't do this one either...

  1. #1
    Member billym's Avatar
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    Integration... can't do this one either...

    I need to find x.
    any help would be great:

    $\displaystyle
    v \frac{dv}{dx} = -\frac{g}{b^2}(v^2 + b^2)
    $
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  2. #2
    Member billym's Avatar
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    Is this right?

    $\displaystyle
    ln(v + 1) = - gx + C
    $

    or this:

    $\displaystyle
    \frac{1}{2} ln(b^2 + v^2) = - \frac{g}{b^2} + C
    $
    Last edited by billym; Mar 26th 2008 at 02:04 AM.
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  3. #3
    Senior Member Peritus's Avatar
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    Quote Originally Posted by billym View Post
    Is this right?

    $\displaystyle
    ln(v + 1) = - gx + C
    $
    first of all you can always check if your answer is correct by plugging it into the ODE. (your answer is wrong)

    let us make the following substitution:

    $\displaystyle
    \begin{gathered}
    z = \frac{1}
    {2}v^2 \hfill \\
    \frac{{dz}}
    {{dx}} = v\frac{{dv}}
    {{dx}} \hfill \\
    \end{gathered}
    $

    thus we get the following equation:

    $\displaystyle
    z' + \frac{{2g}}
    {{b^2 }}z = - g
    $

    This is a first order ODE which can be easily solved...
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  4. #4
    Flow Master
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    Quote Originally Posted by billym View Post
    I need to find x.
    any help would be great:

    $\displaystyle
    v \frac{dv}{dx} = -\frac{g}{b^2}(v^2 + b^2)
    $
    $\displaystyle \Rightarrow \frac{dx}{dv} = -\frac{b^2}{g} \int \frac{v}{v^2 + b} \, dv$.

    Make the substitution $\displaystyle u = v^2 + b$.
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  5. #5
    Member billym's Avatar
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    um...

    $\displaystyle
    x = -\frac{b^2}{2g}ln(v^2 + b)+C
    $
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  6. #6
    Flow Master
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    Quote Originally Posted by billym View Post
    um...

    $\displaystyle
    x = -\frac{b^2}{2g}ln(v^2 + b)+C
    $
    No.

    $\displaystyle x = - \frac{b^2}{2g} \ln |v^2 + b| + C$.

    Note that $\displaystyle \int \frac{1}{u} \, du = \ln |u| + C$ ....
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle \Rightarrow \frac{dx}{dv} = -\frac{b^2}{g} \int \frac{v}{v^2 + b} \, dv$.
    (chuckles) Are you sure about that? I think the LHS is just $\displaystyle \int dx$.

    -Dan
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  8. #8
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    Quote Originally Posted by topsquark View Post
    (chuckles) Are you sure about that? I think the LHS is just $\displaystyle \int dx$.

    -Dan
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  9. #9
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    I think it is all a bit confusing to the original poster. The equation is one that can be separated and thus written as:

    $\displaystyle \int\frac{v \cdot dv}{v^2 + b^2}=-\frac{g}{b^2}\int dx$

    This should get the original poster a step further.
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