How can I prove the solution for this problem involving integration?

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SUMMARY

The discussion centers on proving the solution to the differential equation \ddot{y}+g(t,y)=0 with initial conditions y(0)=y_{0} and \dot{y}(0)=z_{0}. The proposed solution is y(t)=y_{0}+tz_{0}-\int_{0}^{t}(t-s)g(s,y(s))ds. The key point of confusion is the transformation of the double integral, specifically the equivalence of \int_{0}^{t}\{\int_{0}^{s}g(\tau,y(\tau))d\tau\}ds and \int_{0}^{t}\{\int_{\tau}^{t}ds\}g(\tau,y(\tau))d\tau, which is clarified through Fubini's theorem regarding the switch of integration order.

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ReyChiquito
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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.
 
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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.
 
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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