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  • Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 57 maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 57 maths textbook solution

Answers (1)

Answer: \frac{2 \log 2}{\log \frac{11}{10}}

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

Given: In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. We have to find, in how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.

Solution: Let the count of bacteria be N

        As the rate of growth of bacteria is proportional to the no. present

        \begin{aligned} &\therefore \frac{d N}{d t} \alpha N \\\\ &\Rightarrow \frac{d N}{d t}=K N \\\\ &\Rightarrow \frac{d N}{N}=K d t \end{aligned}

          Integrating both sides

        \begin{aligned} &\int \frac{d N}{N}=K \int d t\\\\ &\Rightarrow \log |N|=K t+c \end{aligned}                        .................(1)

        InitiallyN=100000;\; t=0

        By (1)

        \begin{aligned} &\log 100000=K(0)+c \\\\ &\Rightarrow c=\log 100000 \end{aligned}

        Put in (1) we get

        \log N=K t+\log 100000                ...................(2)

        According to given

         When

          \begin{aligned} \mathrm{t}=2, \mathrm{~N} &=10 \% \text { of } 100000+100000 \\ &=10000+100000=110000 \end{aligned}

          \begin{aligned} &\operatorname{By}(2) \log 110000=\mathrm{K}(2)+\log 100000 \\\\\ &\log 110000=K(2)+\log 100000 \\\\ &\Rightarrow \log \frac{110000}{100000}=K(2) \\\\ &{\left[\log m-\log n=\log \frac{m}{n}\right]} \end{aligned}

        \begin{aligned} &{\left[\log m-\log n=\log \frac{m}{n}\right]} \\\\ &\Rightarrow K=\frac{1}{2} \log \frac{11}{10} \\\\ &\therefore \log N=\frac{1}{2} \log \frac{11}{10} t+\log 100000 \end{aligned}            ....................(3)

        Now we have to find in how many hours i.e t1 ; N=200000

        By (3)

        \begin{aligned} &\log 200000=\frac{1}{2}\left(\log \frac{11}{10}\right) t^{1}+\log 100000 \\\\ &\Rightarrow \log \left|\frac{200000}{100000}\right|=\frac{1}{2} \log \left(\frac{11}{10}\right) t^{1} \\\\ &\Rightarrow \log 2=\frac{1}{2} \log \left(\frac{11}{10}\right) t^{1} \end{aligned}

        \begin{aligned} &\Rightarrow 2 \log 2=\log \left(\frac{11}{10}\right) t^{1} \\\\ &\Rightarrow \frac{2 \log 2}{\log \left(\frac{11}{10}\right)}=t^{1} \\\\ &\Rightarrow t^{1}=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \end{aligned}

 

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