Chapter 6 - Differential Equations - 6.3 Exercises - Page 421: 23

$$P=P_0e^{kt}$$

$dP-kPdt=0$ $dP=kPdt$ $\frac{1}{P}dP=kdt$ $\int\frac{1}{P}dP=\int kdt$ $\ln{P}=kt+C$ $e^{\ln{P}}=P=e^{kt+C}=e^Ce^{kt}$ Let $A=e^C$ Then $P=Ae^{kt}$ Applying our initial condition we get $P_0=Ae^0=A$ and so our particular solution becomes $$P=P_0e^{kt}$$