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Thread: differencial equation

  1. #1
    Junior Member cinder's Avatar
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    differencial equation



    Can someone give me a starting point?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cinder View Post


    Can someone give me a starting point?
    Here's the first part:

    $\displaystyle \frac{dV}{dt} = -0.03V$

    $\displaystyle \frac{dV}{V} = -0.03dt$

    $\displaystyle \int_{V_0}^V \frac{dV}{V} = -0.03 \int_0^t~dt$

    $\displaystyle ln(V)|_{V_0}^V = -0.03t|_0^t$

    $\displaystyle ln(V) - ln(V_0) = -0.03t$

    $\displaystyle ln \left ( \frac{V}{V_0} \right ) = -0.03t$

    $\displaystyle V = V_0e^{-0.03t}$

    -Dan
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cinder View Post


    Can someone give me a starting point?
    Here's the second part.

    Dan provided the equation for V(t)

    It is $\displaystyle V(t) = V_0 e^{-0.03t}$

    Now we want t such that V(t) is 25% of it's original value, in other words, we want t when

    $\displaystyle V(t) = 0.25V_0$

    So our equation becomes

    $\displaystyle 0.25V_0 = V_0 e^{-0.03t}$

    $\displaystyle \Rightarrow 0.25 = e^{-0.03t}$

    $\displaystyle \Rightarrow t = - \frac {\ln 0.25}{0.03}$
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  4. #4
    Junior Member cinder's Avatar
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    Thanks for the help, I think I got it!
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