Question

Im not sure how to integrate this one.. what's the trick?
image and latex coming ...

Answer 1

\[E^{-r t} A'[t] - r E^{-r t} A[t] = -p[t] E^{-r t}\]

Answer 2

if it helps I know already that..
A'[t] = r A[t] - p[t]​
p[t] = r A[t] - A'[t]

Answer 3

\[\int\limits\limits_{0}^{\infty} -p(t) E^{-r t} dt = ???\]

Answer 4

\[
\begin{align*}
&\phantom{=}\int_0^\infty e^{-rt}A'(t)-re^{-rt}A(t)\,dt\\
&=[e^{-rt}A(t)]_0^\infty\\
&=-A(0)\\
\end{align*}
\]
\[
\begin{align*}
-A(0)&=\int_0^\infty -p(t)e^{-rt}\,dt\\
A(0)&=\int_0^\infty p(t)e^{-rt}\,dt
\end{align*}
\]

Answer 5

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.

Answer 6

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. ...

Answer 7

The instructions above just differentiated \(e^{-rt}A(t)\). Of course the result of integrating what was differentiated is the thing being differentiated lol!