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Math Help - first order ODE question.

  1. #1
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    first order ODE question.

    I have been given an equation which is first order ODE of the velocity v(t).

    v'(t) = g-(k/m)v(t)

    Where g is 9.8m/s^2, k is a constant of proportionality based on the size and shape of the body falling, and m is the mass.

    I am asked to find the general solution of this DE, satisfying the initial value v(o) = v0(initial velocity).

    I believe that I am supposed to take g-(k/m) to be a constant. Any suggestions as to how i should go about solving this?
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  2. #2
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    Could someone please verify if this is correct. This is the answer I am coming up with.

    v(t) = 9.8m/k - (v(0)+9.8m/k)e^kt/m
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  3. #3
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    Quote Originally Posted by 04viking View Post
    Could someone please verify if this is correct. This is the answer I am coming up with.

    v(t) = 9.8m/k - (v(0)+9.8m/k)e^kt/m
    I get v = \frac{mg - (mg - k v_0) e^{-kt/m}}{k} = \frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}.

    The DE is \frac{dv}{dt} = \frac{mg - kv}{m} \Rightarrow \frac{dt}{dv} = \frac{m}{mg - kv}.

    The DE clearly suggests a terminal velocity v_t = \frac{mg}{k}, which is consistent with my solution but obviously not yours (your solution has v --> +oo as t --> +oo).

    At t = 0 my solution gives v = v_0 whereas your solution gives v = -v_0 ....
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  4. #4
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    Thank you very much for the help.
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  5. #5
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    I have run into a small problem. I am given that v(t) = s'(t) and to solve for s(t) when s(0)=0. I cannot seem to solve it. Could someone please show me a few steps in the right direction?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I get v = \frac{mg - (mg - k v_0) e^{-kt/m}}{k} = \frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}.

    The DE is \frac{dv}{dt} = \frac{mg - kv}{m} \Rightarrow \frac{dt}{dv} = \frac{m}{mg - kv}.

    The DE clearly suggests a terminal velocity v_t = \frac{mg}{k}, which is consistent with my solution but obviously not yours (your solution has v --> +oo as t --> +oo).

    At t = 0 my solution gives v = v_0 whereas your solution gives v = -v_0 ....
    Quote Originally Posted by 04viking View Post
    I have run into a small problem. I am given that v(t) = s'(t) and to solve for s(t) when s(0)=0. I cannot seem to solve it. Could someone please show me a few steps in the right direction?
    Set up the differential equation.

    mr fantastic has found v for you. Since v=\frac{\,ds}{\,dt}, it can be easily determined that \frac{\,ds}{\,dt}=\frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}

    This implies that \int\,ds=\int\left[\frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}\right]\,dt\implies s(t)=\dots

    Then apply the initial condition s(0)=0 to find the value of the constant of integration C.

    --Chris
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