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Thread: v' = av - b(v^2)

  1. #1
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    v' = av - b(v^2)

    x=x(t)

    I need to solve x''=ax'-b(x')^2 to find x' in terms of t.

    I rewrote it using x' = v:

    v' = av - b(v^2)

    From here I'm stuck. It's mainly the v^2 term that's putting me off.

    Any help or hints would be much appreciated.
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  2. #2
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    Re: v' = av - b(v^2)

    I need to solve x''=ax'-b(x')^2 to find x' in terms of t.
    assuming $a$ and $b$ are constants ...

    $\displaystyle \frac{dv}{dt} = av - bv^2$

    $\displaystyle \frac{dv}{av-bv^2} = dt$

    $\displaystyle \frac{dv}{v(a-bv)} = dt$

    partial fraction decomposition ...

    $\displaystyle \frac{1}{a}\left( \frac{1}{v} + \frac{b}{a-bv}\right) \, dv = dt$

    $\displaystyle \left( \frac{1}{v} + \frac{b}{a-bv}\right) \, dv = a \, dt$
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