#### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 56 maths

Answer: $1648$

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

Given: In a bank principal increases increases at the rate of 5% per year.

An amount of Rs.1000 is deposited with this bank, how much will it worth after 10 years.(e0.5 =1.648)

Solution: Let P be the Principal

As Principal increases at the rate of 5% w.r.t t

\begin{aligned} &\therefore \frac{d P}{d t}=\frac{5}{100} \times P \\\\ &\Rightarrow \frac{d P}{P}=\frac{1}{20} d t \end{aligned}

Integrating both sides

$\int \frac{d P}{P}=\frac{1}{20} \int 1 d t$                ................(1)

Now initially P0=1000; after 10 years P= P10 also initially t =0 and after 10 years t = 10

By (1)

\begin{aligned} &\int_{P_{0}}^{P_{0}} \frac{d P}{P}=\frac{1}{20} \int_{0}^{+} 1 d t \\\\ &\Rightarrow[\log |P|]_{1000}^{P_{0}}=\frac{1}{20}[t]_{0}^{10} \\\\ &\Rightarrow \log P_{10}-\log 1000=\frac{1}{20}(10-0) \end{aligned}

\begin{aligned} &\Rightarrow\left[\log \left|\frac{P_{10}}{1000}\right|\right]=\frac{1}{2} \\\\ &{\left[\log m-\log n=\log \frac{m}{n}\right]} \\\\ &\Rightarrow\left|\frac{P_{10}}{1000}\right|=e^{\frac{1}{2}} \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} \log _{a} e=x \\ a=e^{x} \end{array}\right]} \\\\ &\Rightarrow P_{10}=1000(1.648) \\\\ &{\left[e^{\frac{1}{2}}=1.648\right]} \\\\ &\Rightarrow P_{10}=1648 \end{aligned}

Principal after 10 years = 1648