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Derivative of a Convolution

  1. Sep 2, 2012 #1

    I want to verify that the form of a particular solution satisfies the following ODE:

    v' + (b/m)v = u/m


    vpart= ∫e-(b/m)(t-r) (u(r)/m) dr

    where the limits are from 0 to t

    So I tried to differentiate v with respect to t, in order to substitute it back into the equation. But, how do you do that when the integral is with respect to r? Is there a need to change variables? How can you do this?

  2. jcsd
  3. Sep 2, 2012 #2


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    v(t) = ∫tf(t, r).dr
    v(t+δt)= ∫t+δtf(t+δt, r).dr
    = ∫tf(t+δt, r).dr + ∫tt+δtf(t+δt, r).dr
    So v' = ∫t(d/dt)f(t, r).dr + f(t, t)
  4. Nov 8, 2012 #3
    Sorry I'm not familiar with your method. I don't understand why you substitute "t+δt" for t. What approach are you using here? Could you elaborate or direct me to some further reading?

    Cheers :)
  5. Nov 8, 2012 #4


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    The equation you posted for v(t) is generic - i.e. it's true for all t. So it's true both for a given t and for a later time t+δt. So you can write a second equation substituting t+δt for t consistently. Taking the difference, diving by δt, then letting δt tend to zero gives you v'. That is the standard process of differentiation.
  6. Nov 8, 2012 #5
    Oh I see what you mean. Thanks for the clarification. I'm just not use to this notation :)
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