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Solving a ODE

  1. Feb 2, 2006 #1
    can you help me olve this (simple?) differential equation
    [tex] m \ddot{y} = -mg - \beta \dot{y} [/tex]
    integrate once and i get
    [tex] \dot{y} = -gt - \frac{\beta y}{m} + C [/tex]
    also can be written as
    [tex] \frac{dy}{dt} = -gt - \frac{\beta y}{m} + C [/tex]
    both are equivalent

    basically trying to get the y on one side adn the t on the other side. HJva tried many ways but cant isolate the two. Any suggestions?
    Last edited: Feb 2, 2006
  2. jcsd
  3. Feb 2, 2006 #2


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    Staff: Mentor

    Have you tried the approach where you assume a solution for y(t), differentiate it twice and work out the equation constants....? What would be a typical function that could work with this approach...?
  4. Feb 2, 2006 #3
    First of all let v = dy/dt, then you have
    [itex]m \frac{dv}{dt} = -mg - \beta v[/itex]
    From which

    [itex] \frac{dv}{g + \frac{\beta}{m} v} = -dt [/itex]

    Integrating it now gives

    [itex] \frac{m}{\beta} \ln(g + \frac{\beta}{m} v) = -t + C_1 [/itex]

    Solving for v gives:

    [itex] v = C e^{-\frac{\beta}{m}t} - \frac{mg}{\beta} [/itex]

    Where C is some new constant which is I think [itex] C = \frac{m}{\beta}e^{\frac{\beta}{m} C_1} [/itex]. But it doesn't matter though.

    Now remember that v = dy/dt. Thus integrating the last equation will give you y(t). (there will be two constants then)
    Last edited: Feb 2, 2006
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