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

1. anonymous

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. 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*}

6. 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.

7. 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. ...

8. 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!