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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


          \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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