TITLE: Exponential growth/decay formula: what happened to the other constant of integration?
QUESTION [5 upvotes]: 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?

REPLY [8 votes]: 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.

REPLY [5 votes]: Well, notice that:
$$\ln\left|\text{P}\left(t\right)\right|+\text{C}_1=\text{k}\cdot t+\text{C}_2\tag1$$
Getting $\text{C}_1$ on the other side gives:
$$\ln\left|\text{P}\left(t\right)\right|=\text{k}\cdot t+\text{C}_2-\text{C}_1\tag2$$
But $\text{C}_2-\text{C}_1$ is another constant, so:
$$\ln\left|\text{P}\left(t\right)\right|=\text{k}\cdot t+\text{C}\tag3$$