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Differential equation - not sure where to start

  1. Aug 21, 2008 #1
    1. The problem statement, all variables and given/known data

    I need to solve for the voltage across a capacitor in a resistor-capacitor charging equation (eg. this circuit), except that the resistance varies inversely with the voltage across the capacitor. With a constant R, I'm happy solving the first order differential equation, but when R(t)=1/(a+b(Vc(t))), I don't know where to start.

    2. Relevant equations

    [tex]V_{in} = V_{r} + V_{c}[/tex]

    [tex]V_{in} = i(t).R(t) + \frac{1}{C}\int_{0}^{t}i(\tau)d\tau[/tex]

    [tex]R(t) = \frac{1}{a+bV_{c}(t)}[/tex]

    3. The attempt at a solution

    [tex] V_{in} = \frac{i(t)}{a+bV_{c}(t)} + \frac{1}{C}\int_{0}^{t}i(\tau)d\tau[/tex]

    [tex] V_{in} = \frac{i(t)}{a+\frac{b}{C}\int_{0}^{t}i(\tau)d\tau} + \frac{1}{C}\int_{0}^{t}i(\tau)d\tau[/tex]

    [tex] V_{in}.a + \frac{b.V_{in}}{C}\int_{0}^{t}i(\tau)d\tau = i(t) + (\frac{1}{C}\int_{0}^{t}i(\tau)d\tau)^{2}[/tex]

    (differentiate wrt t, use chain rule for the last term)

    [tex] 0 + \frac{b.V_{in}}{C}i(t) = \frac{di(t)}{dt} + \frac{2.i(t)}{C}\int_{0}^{t}i(\tau)d\tau)[/tex]

    I'm a bit stuck with where to go here - I'm not sure how to get rid of the integral in the last term, or even whether I've done the right thing so far. Any hints would be greatly appreciated,

  2. jcsd
  3. Aug 21, 2008 #2


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    Homework Helper

    It looks easier if you are to work in terms of q(t) instead of i(t) since you won't have to deal with the integral of i(t). If you write everything out in terms of q and throw the dq/dt to one side and everything else on the left, the DE is recognisable as a Riccati differential equation.

    On the other hand you could make use of the Laplace transform to work with i(t) but the term in R(t) looks pretty complicated to transform.
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