- #1
ReyChiquito
- 120
- 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.
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.