#### Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 55

Answer: $6.931$

Hint: Separate the terms of x and y and then integrate them.

Given: In a bank principal increases at the rate of r% per year. We have to find the value of r if Rs.100 double itself in 10 years (log 2=0.6931)

Solution: Let Principal = P and rate = r

As principal increases at the rate of r% w.r.t time i.e. per year.

\begin{aligned} &\therefore \frac{d P}{d t}=\frac{r}{100} \times P \\\\ &\Rightarrow \frac{d P}{P}=\frac{r}{100} d t \end{aligned}

Integrating both sides

$\Rightarrow \log P=\frac{r}{100} t+c$            ..................(1)

Suppose initially $t = 0,P = P_{0}$

\begin{aligned} &\Rightarrow \log P_{0}=\frac{r}{100} 0+c \\\\ &\Rightarrow c=\log P_{0} \end{aligned}

Put in (1)

\begin{aligned} &\Rightarrow \log P=\frac{r}{100} t+\log P_{0} \\\\ &\log P-\log P_{0}=\frac{r}{100} t \end{aligned}                        ..............(2)

According to given

When $t = 10, P = 2 * 100, P_{0}=100$

By (2)

\begin{aligned} &\log 200-\log 100=\frac{r}{100} 10 \\\\ &\Rightarrow \log \left(\frac{200}{100}\right)=\frac{r}{10} \\\\ &\Rightarrow \log 2=\frac{r}{10} \Rightarrow 10 \log _{e} 2=r \\ &\Rightarrow r=10 \times 0.6931=6.931 \\ &\Rightarrow r=6.931 \end{aligned}