#### Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 9 maths textbook solution

Given:  Compound interest = 8% per annum,

To find: The percentage increase over one year.

Hint: Use compound interest i.e. $\frac{d P}{d t}=\frac{p r}{100}$

Solution: Let P be the principal

\begin{aligned} &=\frac{d P}{d t}=\frac{p r}{100} \\\\ &=\frac{d P}{P}=\frac{r}{100} d t \end{aligned}

Integrating on both sides we get,

\begin{aligned} &=>\int \frac{d P}{P}=\int \frac{r}{100} d t \\\\ &=>\log P=\frac{r}{100} \times t+C \ldots(i) \end{aligned}    [Where C is integral constant]

At t=0, we have initial principal P=P0

\begin{aligned} &=>\log \left(P_{0}\right)=0+C \\\\ &=>C=\log P_{0} \end{aligned}

Now substituting value of C in equation (i)

\begin{aligned} &=>\log P=\frac{r}{100} t+\log P_{0} \\\\ &=>\log P-\log P_{0}=\frac{r t}{100} \\\\ &=>\log \left(\frac{p}{p_{0}}\right)=\frac{r t}{100} \ldots(i i) \end{aligned}

For t=1 and r=8% in equation (ii)

\begin{aligned} &=>\log \left(\frac{p}{p_{0}}\right)=\frac{8}{100} \\\\ &=>\log \left(\frac{p}{p_{0}}\right)=0.08 \end{aligned}

\begin{aligned} &=>\frac{P}{P_{0}}=e^{0.08} \\\\ &=>\frac{P}{P_{0}}=1.0833 \quad\left[e^{0.08}=1.0833\right] \end{aligned}

Subtracting 1 on both sides we get

\begin{aligned} =&>\frac{P}{P_{0}}-1=1.0833-1 \\\\ \end{aligned}

$=\frac{P-P_{0}}{P_{0}} \times 100$

\begin{aligned} &=0.0833 \times 100 \\\\ &=8.33 \% \end{aligned}

Hence, amount increases by 8.33% in one year.