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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Ex 9.4, 22 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? Let the number of bacteria at time t be y Given that rate of growth of bacteria is proportional to the number present ๐‘‘๐‘ฆ/๐‘‘๐‘ก โˆ y ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ky ๐‘‘๐‘ฆ/๐‘ฆ = kdt Integrating both sides โˆซ1โ–’ใ€–๐‘‘๐‘ฆ/๐‘ฆ=๐‘˜ใ€— โˆซ1โ–’๐‘‘๐‘ก log y = kt + C Now, according to question 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 Putting t = 0 and y = 1,00,000 in (1) log 1,00,000 = k ร— 0 + C C = log 1,00,000 โ€ฆ(1) Putting value of C in (1) log y = kt + C log y = kt + log 1,00,000 Now, Putting t = 2 and y = 1,00,000 in (2) log 1,10,000 = 2k + log 1,00,000 log 1,10,000 โˆ’ log 1,00,000 = 2k log (1,10,000/1,00,000) = 2k 1/2 log (11/10) = k โ€ฆ(2) Putting value of k in (2) log y = kt + log 1,00,000 log y = 1/2 log (11/10) t + log 1,00,000 Now, If Bacterial = 2,00,000, we have to find t Putting y = 2,00,000 in (3) log 2,00,000 = 1/2 log (11/10) t + log (1,00,000) log 2,00,000 โˆ’ log 1,00,000 = 1/2 log (11/10) t log (2,00,000/1,00,000) = 1/2 log (11/10) t log 2 = 1/2 log (11/10) t t = (๐Ÿ ๐ฅ๐จ๐ โก๐Ÿ)/๐ฅ๐จ๐ โกใ€– (๐Ÿ๐Ÿ/๐Ÿ๐ŸŽ)ใ€— โ€ฆ(3)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.