Chapter 6 - Differential Equations - Review Exercises - Page 431: 24

$y=5e^{-0.68t}$

$(0, 5)$ lies on the curve. Thus, $$5 = Ce^{k.0}$$ Or, $$C=5$$ $(5, \frac{1}{6})$ lies on the curve. Thus, $$\frac{1}{6} = 5e^{5k}$$ Or, $$e^{5k}=\frac{1}{30}$$ $$k = \frac{\ln(1/30)}{5} = -0.68$$ So, the function is $$y=5e^{-0.68t}$$