The standard equation for exponential growth and decay starts and is derived like this:
$$ {dP\over dt}=kP$$
$$ {dP\over P}=kdt$$
$$ \int{dP\over P}=\int kdt$$
$$ \color{red}{\ln |P|}=kt+C$$
I don't understand the left hand side at this point, isn't $\int{1\over x}dx = \ln |x| +C$? Where did the constant of integration from the left integral go?
Answer
When you integrate both sides, each has a constant - you'd get, for constants $A,B$:
$$ \int{dP\over P}=\int kdt \implies \ln|P|+A = kt+B$$
Well, we can subtract $A$ from both sides and define a constant $C = B-A$; then
$$\ln|P|+A = kt+B \implies \ln|P|=kt+B-A=kt+C$$
This combination of constants is often implicit in solving differential equations - you'll integrate on two sides and then just combine the constants on whichever side of the equation is more convenient.
No comments:
Post a Comment