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1.1.2 Linear equations

The most general first order 1-D linear ODE is of the form:

$\displaystyle \frac{dv}{dt}= -\alpha (t) v(t) + I(t), \qquad v(0) = v_0$ (1)

We derived solutions for 5 different cases:
  1. $ \alpha (t) = \alpha (= const)$, $ I(t) = 0$. We showed that the solution in this case is given by:

    $\displaystyle v(t) = v_0 \mathrm{e}^{-\alpha t} $

    We studied two different cases: exponential decay $ (\alpha > 0)$ using the RC circuit example and exponential growth $ (\alpha <0)$ using the credit card example.
  2. $ \alpha (t) = \alpha (t)$, $ I(t) = 0$. We showed that the solution in this case is given by:

    $\displaystyle v(t) = v_0 \mathrm{e}^{-\int_0^t\alpha (s)ds} $

    We used a credit card example with variable APR to demonstrate and get intuition for the structure of the solution.
  3. $ \alpha (t) = \alpha (= const)$, $ I(t) = I (=const)$. We showed that the solution in this case is given by:

    $\displaystyle v(t) = v_0 \mathrm{e}^{-\alpha t} + I \int_0^t\mathrm{e}^{-\alpha (t-s)} ds $

  4. $ \alpha (t) = \alpha (= const)$, $ I(t) = I(t)$. We showed that the solution in this case is given by:

    $\displaystyle v(t) = v_0 \mathrm{e}^{-\alpha t} + \int_0^tI(s) \mathrm{e}^{-\alpha (t-s)} ds $

  5. $ \alpha (t) = \alpha (t)$, $ I(t) = I(t)$. We showed that the solution in the most general linear case is given by:

    $\displaystyle v(t) = v_0 \mathrm{e}^{-\int_0^t\alpha (s)ds} + \int_0^tI(s) \mathrm{e}^{-\int_s^t \alpha (u) du} ds $


next up previous contents
Next: 1.1.3 Some non linear Up: 1.1 One dimensional differential Previous: 1.1.1 Differential equations and
Eran Borenstein
2007-05-04