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Question Number 210025 by professorleiciano last updated on 29/Jul/24

Commented by mr W last updated on 29/Jul/24

$(b)$ $\frac{d C}{i C + k} = d t$ $\int_{C_{0}}^{C} \frac{d C}{i C + k} = \int_{0}^{t} d t$ $\frac{1}{i} [\ln (i C + k) - \ln ({i C}_{0} + k)] = t$ $\ln \frac{i C + k}{{i C}_{0} + k} = i t$ $\frac{i C + k}{{i C}_{0} + k} = e^{i t}$ $\Rightarrow C = (C_{0} + \frac{k}{i}) e^{i t} - \frac{k}{i}$ $(c)$ $f o r k = 0 a n d i = 0 . 75 % :$ $\frac{C}{C_{0}} = e^{i t} \Rightarrow 2 = e^{0.0075 t} \Rightarrow t = \frac{\ln 2}{0.0075} \approx 93$ $i . e . 93 m o n t h e s a r e r e q u i r e d .$
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