Chapter 9 - Differential Equations - 9.3 Separable Equations - 9.3 Exercises - Page 646: 43

$C(t) = \frac{-e^{-kt}}{c'} + \frac{r}{k}$ {where $c'$ = ${kc^{-k}}$}

$\frac{dC}{dt} = r-kC$ $\frac{dC}{r-kC} = dt$ $\frac{ln(r-kC(t))}{-k} + lnc= t$ $\ln(r-kC(t)) + ln(c^{-k})= -kt$ $(r-kC(t))(c^{-k}) = e^{-kt}$ $(r-kC(t)) = \frac{e^{-kt}}{c^{-k}}$ $kC(t) = r-\frac{e^{-kt}}{c^{-k}}$ $C(t) = \frac{-e^{-kt}}{c'} + \frac{r}{k}$ {where $c'$ = ${kc^{-k}}$}