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Singular solution for a differential equation?

  1. Sep 27, 2011 #1
    Hey, this is my first time on the forums so sorry if I do something wrong...
    Let c'(t) = f(c) = [itex]\frac{(kr + P - c(t)r)}{V}[/itex]. Determine the singular solution of Vc'(t)

    k, P, r, and V are all constant

    I'm not quite sure if this is the correct way to find the singular solution but I used this equation:
    Integrating Factor Method:
    To solve a linear DE
    y' + x(t)y = f(t)

    where x and f are continuous on the domain I:
    u(t) = e[itex]^\int{x(t) dt}[/itex]

    u(t)[y' + x(t)y] = f(t)u(t)

    [itex]\frac{d}{dt}[/itex][u(t)y(t)] = f(t)u(t)

    u(t)y(t) = [itex]\int{f(t)u(t )dt}[/itex] + j

    where j is a constant

    y = ([itex]\frac{\int{f(t)u(t) dt} + j)}{u(t)}[/itex]

    f'(c) = -[itex]\frac{c'(t)r}{V}[/itex]

    f'(c) = -[itex]\frac{f(c)r}{V}[/itex]

    f'(c) + [itex]\frac{f(c)r}{V}[/itex] = 0

    let x = [itex]\frac{r}{V}[/itex]

    u = e[itex]^{\int{xdt}}[/itex]

    u = e[itex]^{\frac{d}{dt}}[/itex]

    [itex]\frac{d}{dt}[/itex][e[itex]^{\frac{rt}{V}}[/itex]f(c)] = 0

    e[itex]^{\frac{rt}{V}}[/itex]f(c) = j

    f(c) = [itex]\frac{j}{e^\frac{rt}{V}}[/itex]

    There is no singular solution

    Now that can't be right can it? I'm completely confused and now I'm second guessing myself everywhere. Any help?
  2. jcsd
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