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General solution of double integrals

  1. Nov 19, 2012 #1
    I have a differential equation of the form and I want to solve it using calculus, as opposed to using a differential equation method.
    [itex]\frac{d^2v}{dt}[/itex] = [itex]\alpha[/itex]

    where [itex]v[/itex] is a function of [itex]t[/itex] i.e., [itex]v(t)[/itex]
    and [itex]\alpha[/itex] is some constant.

    How do I solve for [itex]v(t)[/itex] if the time ranges from [itex]t_0[/itex] to [itex]t [/itex] such that [itex]v'(t_0)[/itex] and [itex]v(t_0)[/itex] are the lower bounds of my double integrals?

    Do I setup a double integral, such as the following?

    [itex]\int_{v(t_0)}^{v(t)}[/itex] [itex]\int_{v'(t_0)}^{v(t)} dvdv [/itex] = [itex]\alpha[/itex][itex]\int_{t_0}^{t}[/itex] [itex]\int_{t_0}^{t} dτ dτ [/itex]

    I get the following solution from evaluating that double integral:
    [itex]v(t) = \frac{\alpha}{2}{(t-t_0})^2 +v'(t_0)(t-t_0) + v(t_0)[/itex]
    and I was wondering if the solution is correct?
     
  2. jcsd
  3. Nov 19, 2012 #2

    Why wonder? Derive twice what you got and see if you get back your differential equation!

    DonAntonio
     
  4. Nov 19, 2012 #3
    I did, but I want to check if the procedure is correct.
     
  5. Nov 21, 2012 #4

    haruspex

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    The LHS makes no sense to me.
    [itex]v''(u) = f(u)[/itex]
    [itex]\left[v'(u)\right]_{u=t_0}^w = \int_{u=t_0}^w f(u) du[/itex]
    [itex]v'(w) - v'(t_0) = \int_{u=t_0}^w f(u) du[/itex]
    [itex]\left[v(w)\right]_{w=t_0}^t - (t-t_0)v'(t_0) = \int_{w=t_0}^t \int_{u=t_0}^w f(u) du dw[/itex]
    [itex]v(t) = v(t_0) + (t-t_0)v'(t_0) + \int_{w=t_0}^t \int_{u=t_0}^w f(u) du dw[/itex]
     
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