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Q regarding Integration

  1. Aug 10, 2004 #1
    Im trying to understand a "simple" identity.

    I have the problem [tex]\ddot{y}+g(t,y)=0[/tex]
    with [tex]y(0)=y_{0}[/tex] and [tex]\dot{y}(0)=z_{0}[/tex]

    What i need to prove is that the solution can be expresed in the form
    [tex]y(t)=y_{0}+tz_{0}-\int_{0}^{t}(t-s)g(s,y(s))ds[/tex]

    Integrating twice is clear that
    [tex]y(t)=y_{0}+tz_{0}-\int_{0}^{t}\{\int_{0}^{s}g(\tau,y(\tau))d\tau\}ds[/tex]

    Now the book goes
    [tex]\int_{0}^{t}\{\int_{0}^{s}g(\tau,y(\tau))d\tau\}ds=\int_{0}^{t}\{\int_{\tau}^{t}ds\}g(\tau,y(\tau))d\tau[/tex]

    ???????????????????? How do i prove that? should i do a change of variable? use fubini?

    Im lost, pls help.
     
  2. jcsd
  3. Aug 10, 2004 #2

    mathman

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    It is not any change of variables. If you look at the domain of integration in the (s,tau) plane, you will see it is a triangle. All the book is describing is how the triangle is expressed when you switch the order of integration. Fubini's theorem says you can do the switch.
     
  4. Aug 10, 2004 #3
    Ohhhhhhhhhhhhh.... i know i should have named the tread "stupid Q about integration".
    Thx a lot dood. I sure need to review my calc notes.
     
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