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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
WiredFaculty
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
WiredFaculty

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