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Math Help - 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:

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

    <br />
ln(v + 1) = - gx + C<br />

    or this:

    <br />
\frac{1}{2} ln(b^2 + v^2) = - \frac{g}{b^2} + C<br />
    Last edited by billym; March 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?

    <br />
ln(v + 1) = - gx + C<br />
    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:

    <br />
\begin{gathered}<br />
  z = \frac{1}<br />
{2}v^2  \hfill \\<br />
  \frac{{dz}}<br />
{{dx}} = v\frac{{dv}}<br />
{{dx}} \hfill \\ <br />
\end{gathered} <br />

    thus we get the following equation:

    <br />
z' + \frac{{2g}}<br />
{{b^2 }}z =  - g<br />

    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:

    <br />
v \frac{dv}{dx} = -\frac{g}{b^2}(v^2 + b^2)<br />
    \Rightarrow \frac{dx}{dv} = -\frac{b^2}{g} \int \frac{v}{v^2 + b} \, dv.

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

    <br />
x = -\frac{b^2}{2g}ln(v^2 + b)+C<br />
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  6. #6
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    Quote Originally Posted by billym View Post
    um...

    <br />
x = -\frac{b^2}{2g}ln(v^2 + b)+C<br />
    No.

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

    Note that \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
    \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 \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 \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:

    \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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