anonymous
  • anonymous
Im not sure how to integrate this one.. what's the trick? image and latex coming ...
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
\[E^{-r t} A'[t] - r E^{-r t} A[t] = -p[t] E^{-r t}\]
anonymous
  • anonymous
if it helps I know already that.. A'[t] = r A[t] - p[t]‚Äč p[t] = r A[t] - A'[t]

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anonymous
  • anonymous
\[\int\limits\limits_{0}^{\infty} -p(t) E^{-r t} dt = ???\]
thomas5267
  • thomas5267
\[ \begin{align*} &\phantom{=}\int_0^\infty e^{-rt}A'(t)-re^{-rt}A(t)\,dt\\ &=[e^{-rt}A(t)]_0^\infty\\ &=-A(0)\\ \end{align*} \] \[ \begin{align*} -A(0)&=\int_0^\infty -p(t)e^{-rt}\,dt\\ A(0)&=\int_0^\infty p(t)e^{-rt}\,dt \end{align*} \]
thomas5267
  • thomas5267
If \(\lim_{x\to \infty}A(t)=c\in \mathbb{R}\) then \([e^{-rt}A(t)]_0^\infty=-A(0)\). If \(\lim_{x\to \infty}A(t)=\pm\infty\) then we have some trouble.
anonymous
  • anonymous
Thank you.. thomas.. I was fishing for ages for a function that would give me the antiderivative to p(t) e^(-rt) so I could use the Fundamental Formula but I couldn't see the connection that A(t) e^(-rt) was the anti-derivative to r A(t) e^(-rt) - A'(t) e^(-rt) *head slap*! awesome. ...
thomas5267
  • thomas5267
The instructions above just differentiated \(e^{-rt}A(t)\). Of course the result of integrating what was differentiated is the thing being differentiated lol!

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