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

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

Answers (1)

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

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