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Differential Equations

Question
CBSEENMA12033087

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solution
Let y denote the bacteria at any time t.
 From the given condition,
                            dy over dt space equals space straight k space straight y comma space where k is the constant of proportionality.
Separting the variables and integrating, we get,
                     integral 1 over straight y dy space equals space straight k integral dt
therefore space space space space space space space log space straight y space equals space straight k space straight t space space plus space straight c                                ...(1)
Let straight y subscript 0 be bacteria at time t = 0
therefore space space space space space space space space space space space log space straight y subscript 0 space equals space 0 plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space log space straight y subscript 0
therefore space space space space from space left parenthesis 1 right parenthesis comma space log space straight y space equals space straight k space straight t space plus space log space straight y subscript 0 space space space space space space space space space space space space space space rightwards double arrow space space log space open parentheses straight y over straight y subscript 0 close parentheses space equals space straight k space straight t space space space space space... left parenthesis 2 right parenthesis
Now,                          straight y space equals space straight y subscript 0 plus 10 over 100 straight y subscript 0 space equals space 11 over 10 straight y subscript 0 space space space when space straight t space equals space 2
therefore space space space from space left parenthesis 2 right parenthesis comma space space log open parentheses fraction numerator 11 space straight y subscript 0 over denominator 10 space straight y subscript 0 end fraction close parentheses space equals space 2 space straight k space space rightwards double arrow space space space straight k space equals space 1 half log space open parentheses 11 over 10 close parentheses
therefore space space from space left parenthesis 2 right parenthesis comma space space log space open parentheses straight y over straight y subscript 0 close parentheses space equals space 1 half log space open parentheses 11 over 10 close parentheses. space straight t subscript 1                         ...(3)
Let the bacteria be 2,00,000 from 1,00,000 in straight t subscript 1 hours
i.e.                      straight y space equals space 2 space straight y subscript 0 space space space space when space straight t space space equals space straight t subscript 1
therefore space space space space from space left parenthesis 3 right parenthesis comma space space log space open parentheses fraction numerator 2 space straight y subscript 0 over denominator straight y subscript 0 end fraction close parentheses space equals space 1 half log space open parentheses 11 over 10 close parentheses. space straight t subscript 1
therefore space space space space space space log space 2 space equals space 1 half log open parentheses 11 over 10 close parentheses space straight t subscript 1
therefore space space space space space space space space space straight t subscript 1 space equals space fraction numerator 2 space log space 2 over denominator log space open parentheses begin display style 11 over 10 end style close parentheses end fraction space hours. space

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