1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]


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


    [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]


    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook