Question Number 210025 by professorleiciano last updated on 29/Jul/24
Commented by mr W last updated on 29/Jul/24
$$\left({b}\right) \\ $$$$\frac{{dC}}{{iC}+{k}}={dt} \\ $$$$\int_{{C}_{\mathrm{0}} } ^{{C}} \frac{{dC}}{{iC}+{k}}=\int_{\mathrm{0}} ^{{t}} {dt} \\ $$$$\frac{\mathrm{1}}{{i}}\left[\mathrm{ln}\:\left({iC}+{k}\right)−\mathrm{ln}\:\left({iC}_{\mathrm{0}} +{k}\right)\right]={t} \\ $$$$\mathrm{ln}\:\frac{{iC}+{k}}{{iC}_{\mathrm{0}} +{k}}={it} \\ $$$$\frac{{iC}+{k}}{{iC}_{\mathrm{0}} +{k}}={e}^{{it}} \\ $$$$\Rightarrow{C}=\left({C}_{\mathrm{0}} +\frac{{k}}{{i}}\right){e}^{{it}} −\frac{{k}}{{i}} \\ $$$$\left({c}\right) \\ $$$${for}\:{k}=\mathrm{0}\:{and}\:{i}=\mathrm{0}.\mathrm{75\%}: \\ $$$$\frac{{C}}{{C}_{\mathrm{0}} }={e}^{{it}} \:\Rightarrow\mathrm{2}={e}^{\mathrm{0}.\mathrm{0075}{t}} \:\Rightarrow{t}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{0}.\mathrm{0075}}\approx\mathrm{93} \\ $$$${i}.{e}.\:\mathrm{93}\:{monthes}\:{are}\:{required}. \\ $$