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1st Order Differential; bit confused.

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A raindrop falls from rest through mist. Its velocity vms−1 vertically downward, at time t seconds after it starts to fall is modelled by the differential equation

    [itex] (1+t)\frac{dv}{dt} + 3v = (1+t)g-6 [/itex]

    Solve the differential equation to show that

    [itex]v=\frac{g}{4}(1+t)-2+(2-\frac{g}{4})(1+t)^{-3}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    [itex] (1+t)\frac{dv}{dt} + 3v = (1+t)g-6 [/itex]

    [itex] \frac{dv}{dt} + \frac{3v}{1+t} = g-\frac{6}{(1+t)} [/itex]


    Let I be the integration factor

    [itex] I = e^{\int{p(x)}dx} [/itex]
    [itex] p(x)=\frac{3}{1+t} [/itex]
    [itex] I = (1+t)^3 [/itex]

    [itex](1+t)^3\frac{dv}{dt} + 3v (1+t)^2 = g(1+t)^3-6(1+t)^2[/itex]

    [itex](1+t)^3 v = \int g(1+t)^3-6(1+t)^2 dt [/itex]

    [itex](1+t)^3 v = \frac{g}{4} (1+t)^4 - \frac{6}{3}(1+t)^3 + c[/itex]

    [itex]v = \frac{g}{4} (1+t) - 2 + \frac{c}{(1+t)^3}[/itex]

    Why is [itex]c=2-\frac{g}{4}[/itex]

    EDIT: don't worry now, it's the value of V when t = 0. Oops !
     
  2. jcsd
  3. Feb 6, 2009 #2
    Assuming: all is correct.

    Use this initial condition
    v(0) = 0
    and substitute into your final equation.

    Edit: Didn't see the edit
     
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