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## Homework Statement

I need to solve for the voltage across a capacitor in a resistor-capacitor charging equation (eg. http://sub.allaboutcircuits.com/images/05175.png" [Broken]), 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.

## Homework 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]

## 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,

Mark

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