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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.
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Answer

Let $P_0$ be the intial amount and P be the amount at any time t. we have
$\frac{\text{dP}}{\text{dt}}=\frac{8\text{P}}{100}$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=\frac{2\text{P}}{25}$
$\Rightarrow\frac{\text{dP}}{\text{P}}=\frac{2}{25}\text{dt}$
Intergrating both sides with respect to t, We get
$\log\text{P}=\frac{2}{25}\text{t}+\text{C}\ ...(\text{i})$
Now,
$\therefore \log\text{P}_{0}=0+\text{C}$
$\Rightarrow \text{C}=\log\text{P}_{0}$
Putting the value of C in (i), we get
$\log\text{P}=\frac{2}{25}\text{t}+\log\text{P}_{0}$
$\Rightarrow \log\frac{\text{P}}{\text{P}_{0}}=\frac{2}{25}\text{t}$
$\Rightarrow \text{e}^\frac{2}{25}\text{t}=\frac{\text{P}}{\text{P}_{0}}$
To find the amount after 1 year, we have
$ \text{e}^\frac{2}{25}=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow \text{e}^{0.08}=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow 1.0833=\frac{\text{P}}{\text{P}_{0}}$
$\Rightarrow \text{P}=1.0833\text{P}_{0}$
Percentage increase $=\Big(\frac{\text{P}-\text{P}_{0}}{\text{P}_{0}}\Big)\times100\%$
$=\Big(\frac{1.0833 \text{P}_{0}-\text{P}_{0}}{\text{P}_{0}}\Big)\times100\%$
$=0.0833\times100\%$
$=8.33\%$

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