Chapter 6 - Differential Equations - 6.1 Exercises - Page 406: 96

k=±4

We are given that $y=e^{kt}$ satisfies the differential equation $y''-16y=0$. Therefore, $(e^{kt})''-16(e^{kt})=0$ $k^2(e^{kt})-16(e^{kt})=0$ $(k^2-16)e^{kt}=0$ SInce $e^{kt}$ is never zero, we get $k^2-16=0$ or k=±4