Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 23

In the following answer, I assume that k and P_0 are constants.
Then, the given differential equation can be solved by separation of variables:
dP - kPdt = 0
dP = kPdt
Dividing by P results in
(dP)/P = kdt .
Integrating both sides, we obtain
lnP = kt + C , where C is an arbitrary constant. We can now solve for P(t) by rewriting the natural logarithmic equation as an exponential (with the base e) equation:
P = e^(kt + C) .
So, the general solution of the equation is P(t) = e^(kt + C) . Since the initial condition is P(0) = P_0 , we can find C:
P(0) = e^(0 + C) = e^C = P_0 . Therefore,
P(t) = e^(kt)*e^C = P_0e^(kt)
The particular solution of the equation with the given initial condition is
P(t) = P_0e^(kt)

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