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Linear first order differential equations - just a question

  1. Jun 20, 2005 #1
    I thought the general solution of the linear first order differential equation

    [tex]y ^{\prime} + p(t)y = g(t), \qquad y\left( t_0 \right) = y_0[/tex]

    were

    [tex]y = \frac{1}{\mu (t)} \left[ \int \mu (s) g(s) \: ds + \mathrm{C} \right],[/tex]

    where

    [tex]\mu (t) = \exp \int p(s) \: ds[/tex]

    However, I have found this

    [tex]y = \frac{1}{\mu (t)} \left[ \int _{t_0} ^t \mu (s) g(s) \: ds + y_0 \right],[/tex]

    where

    [tex]\mu (t) = \exp \int _{t_0} ^t p(s) \: ds[/tex]

    Can anybody please clarify that? I don't see why those integral limits should be there. Any help is highly appreciated.
     
    Last edited: Jun 20, 2005
  2. jcsd
  3. Jun 20, 2005 #2
    general solution correct.

    however, in this case, integration limits needed for initial condition:

    [tex] y\left( t_0 \right) = y_0 [/tex]

    that is:

    [tex]\mu (t_0) = \exp \int _{t_0} ^{t_0} p(s) \: ds = \exp(0) = 1 [/tex]

    [tex]y(t_0) = \frac{1}{\mu (t_0)} \left[ \int _{t_0} ^{t_0} \mu (s) g(s) \: ds + y_0 \right] = \frac {1} {1} \left [(0) \ + \ y_0 \right ] \ = \ y_0 [/tex]

    nothing really complicated here
     
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