Chapter 1 - Introduction - 1.3 Classification of Differential Equations - Problems - Page 25: 15

$r = -2$

We must find $r$ for which the differential equation $y'+2y = 0$ has solutions of the form $y = e^{rt}$. So, $y'+2y = 0$, $y' = -2y$, $y'/y = -2$, $\int( y'/y) dt = -\int 2 dt$, $\ln y = -2t+C$, $y = e^{C}e^{-2t}$ $=ce^{-2t}$. Thus, $r = -2$.