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I have the following equation

$$\frac{dG(s,t)}{dt}=r(s-1)G(s,t)$$

which is supposed to yield the following when integrating from 0 to $t$ over $t$ (on both sides);

$$G(s,t)=e^{r(s-1)t}G(0,t)$$

Any help is highly appreciated. :)

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  • $\begingroup$ You are using the fact that $(e^x)'=e^x$. $\endgroup$
    – Jon
    Commented Jul 20, 2018 at 10:11
  • $\begingroup$ First answer this for the differential equation $dy(t)/dt = c\;y(t)$. $\endgroup$
    – GEdgar
    Commented Jul 20, 2018 at 10:31
  • $\begingroup$ GEdgar - yes of course :) $\endgroup$
    – user469216
    Commented Jul 20, 2018 at 10:39

1 Answer 1

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Assuming that $G(s,t)$ is positive $\frac {d \ln G(s,t)} {dt} =\frac {dG(s,t)/dt} {G(s,t)}=r(s-1)$. Integrating we get $\ln G(s,t)=rt(s-1)+c$ where $c$ is a constant. Hence $G(s,t)=e^{rt(s-1)+c}$. Put $s=0$ to get $G(0,t)=e^{c-rt}$. X Can you take it from here?

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