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Differentiating a 1st order ODE: really dumb question

  1. Oct 15, 2012 #1

    cepheid

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    Suppose I have a really simple first order linear ODE like:$$\dot{\omega} = -k\omega$$ where k is some constant, ω(t) is a function of time that I want to solve for, and the overdot denotes the derivative w.r.t. time. This is really easy to solve, and we all know that with the initial condition ω(0) = ω0, the solution is given by ω(t) = ω0exp(-kt). Now, I was thinking. You can find a solution that makes this equation true. If this equation is true, then if I differentiate both sides, then the result should also be true right :uhh: (realizes there may be some sketchy reasoning here). So if I differentiate both sides, I get$$\ddot{\omega} = -k\dot{\omega}$$If the first ODE was true given the solution above, this ODE should also be true given that solution. And that is the case, because for that solution, ##\ddot{\omega} = k^2\omega = -k\dot{\omega}##. Now, if I use the symbol ##\alpha## to denote ##\dot{\omega}## then the equation is $$\dot{\alpha} = -k\alpha$$The solution to *this* equation, assuming an initial condition of α(0) = α0, is, of course, α(t) = α0exp(-kt). Now, if I solve for ω(t) by integrating α(t), still assuming that ω(0) = ω0, then I get:$$\omega(t) - \omega_0 = \int_0^t \alpha_0\exp(-kt^\prime)\,dt^\prime$$ $$\omega(t) = \omega_0 + \frac{\alpha_0}{k}[1 - \exp(-kt)] $$So, this solution doesn't have the same form as the original ω(t). What stupid thing have I overlooked here? :redface:
     
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  3. Oct 15, 2012 #2

    D H

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    What you've overlooked is that [itex]\alpha_0[/itex] is not a free variable. It is instead given by [itex]\alpha_0 = -k\omega_0[/itex]. Substitute this into your second [itex]\omega(t)[/itex] and tada! you recover the first one.
     
  4. Oct 15, 2012 #3

    cepheid

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    Thanks for the help! This occurred to me on the walk home. I kind of wish it had occurred to me before posting, but posting definitely helped me sort things out.
     
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