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

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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}


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