Aug 2016
2
0
Australia
Integrate dP/dt = kP solving for P?
And how do I find k if it is a constant?
 
Nov 2013
263
59
Australia
I think this is correct :/


If $P=e^{kt}$ then $\frac{dP}{dt}=ke^{kt}=kP$
so
If $\frac{dP}{dt}=kP$
then
$\int\;\frac{dP}{dt}\;dt=\int \;kP\;dt= e^{kt}+c$
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$\dfrac{dP}{dt}=kP$

separate variables ...

$\dfrac{dP}{P}=k \, dt$

integrate both sides ...

$\ln|P| = kt+C \implies P= \pm e^{kt+C}= \pm e^C \cdot e^{kt}$

assuming $P > 0$ ...

at $t=0$, $P=P_0 \implies e^C=P_0$, resulting in the equation for natural exponential growth (or decay) ...

$P=P_0e^{kt}$

to solve for the growth (or decay) constant, divide both sides by $P_0$, then take the log of both sides to isolate $k$ ...

$k=\dfrac{\ln\left(\frac{P}{P_0}\right)}{t}$ for a given $(t,P)$ besides $(t_0,P_0)$
 
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