Question Number 210025 by professorleiciano last updated on 29/Jul/24
Commented by mr W last updated on 29/Jul/24
![(b) (dC/(iC+k))=dt ∫_C_0 ^C (dC/(iC+k))=∫_0 ^t dt (1/i)[ln (iC+k)−ln (iC_0 +k)]=t ln ((iC+k)/(iC_0 +k))=it ((iC+k)/(iC_0 +k))=e^(it) ⇒C=(C_0 +(k/i))e^(it) −(k/i) (c) for k=0 and i=0.75%: (C/C_0 )=e^(it) ⇒2=e^(0.0075t) ⇒t=((ln 2)/(0.0075))≈93 i.e. 93 monthes are required.](https://www.tinkutara.com/question/Q210030.png)
$$\left({b}\right) \\ $$$$\frac{{dC}}{{iC}+{k}}={dt} \\ $$$$\int_{{C}_{\mathrm{0}} } ^{{C}} \frac{{dC}}{{iC}+{k}}=\int_{\mathrm{0}} ^{{t}} {dt} \\ $$$$\frac{\mathrm{1}}{{i}}\left[\mathrm{ln}\:\left({iC}+{k}\right)−\mathrm{ln}\:\left({iC}_{\mathrm{0}} +{k}\right)\right]={t} \\ $$$$\mathrm{ln}\:\frac{{iC}+{k}}{{iC}_{\mathrm{0}} +{k}}={it} \\ $$$$\frac{{iC}+{k}}{{iC}_{\mathrm{0}} +{k}}={e}^{{it}} \\ $$$$\Rightarrow{C}=\left({C}_{\mathrm{0}} +\frac{{k}}{{i}}\right){e}^{{it}} −\frac{{k}}{{i}} \\ $$$$\left({c}\right) \\ $$$${for}\:{k}=\mathrm{0}\:{and}\:{i}=\mathrm{0}.\mathrm{75\%}: \\ $$$$\frac{{C}}{{C}_{\mathrm{0}} }={e}^{{it}} \:\Rightarrow\mathrm{2}={e}^{\mathrm{0}.\mathrm{0075}{t}} \:\Rightarrow{t}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{0}.\mathrm{0075}}\approx\mathrm{93} \\ $$$${i}.{e}.\:\mathrm{93}\:{monthes}\:{are}\:{required}. \\ $$