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

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

Answers (1)

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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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