ReyChiquito
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- 1
Im trying to understand a "simple" identity.
I have the problem \ddot{y}+g(t,y)=0
with y(0)=y_{0} and \dot{y}(0)=z_{0}
What i need to prove is that the solution can be expresed in the form
y(t)=y_{0}+tz_{0}-\int_{0}^{t}(t-s)g(s,y(s))ds
Integrating twice is clear that
y(t)=y_{0}+tz_{0}-\int_{0}^{t}\{\int_{0}^{s}g(\tau,y(\tau))d\tau\}ds
Now the book goes
\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
? How do i prove that? should i do a change of variable? use fubini?
Im lost, pls help.
I have the problem \ddot{y}+g(t,y)=0
with y(0)=y_{0} and \dot{y}(0)=z_{0}
What i need to prove is that the solution can be expresed in the form
y(t)=y_{0}+tz_{0}-\int_{0}^{t}(t-s)g(s,y(s))ds
Integrating twice is clear that
y(t)=y_{0}+tz_{0}-\int_{0}^{t}\{\int_{0}^{s}g(\tau,y(\tau))d\tau\}ds
Now the book goes
\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
? How do i prove that? should i do a change of variable? use fubini?
Im lost, pls help.