# Question 3 - Miscellaneous - Chapter 9 Class 12 Differential Equations

Last updated at April 8, 2024 by Teachoo

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 1 (iii) Important

Misc 2 (i)

Misc 2 (ii) Important

Misc 2 (iii)

Misc 2 (iv) Important

Misc 3 Important

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13 (MCQ)

Misc 14 (MCQ) Important

Misc 15 (MCQ)

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Important Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams You are here

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 8, 2024 by Teachoo

Question 3 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25000 in the year 2004 , what will be the population of the village in 2009 ? Let the population at time t be y Given that population increases continuously at rate proportional to the number of its inhabitants present at any time 𝑑𝑦/𝑑𝑡 ∝ y 𝑑𝑦/𝑑𝑡 = ky 𝑑𝑦/𝑦 = k dt Integrating both sides ∫1▒𝑑𝑦/𝑦=𝑘∫1▒𝑑𝑡 log y = kt + C In 1999, t = 0 & y = 20000 log 20000 = k(0) + C C = log 20000 Put value in (1) log y = kt + C log y = kt + log 20000 In 2004, t = 5 &y = 25000 Putting values in (2) In 2004, t = 5 &y = 25000 Putting values in (2) log 25000 = 5k + log 20000 log 25000 − log 20000 = 5k log 25000/20000 = 5k 1/5 log 5/4 = k k = 1/5 log 5/4 (As log a – log b = log 𝑎/𝑏) Putting value of k in (2) log y = kt + log 20000 log y = 𝑡/5 log 5/4 + log 20000 Now, In 2009, t = 10 , we need to find value of y Putting t = 10 in (3) log 𝑦 = 10/5 log 5/4 + log 20000 log y − log 20000 = 2 log 5/4 log 𝑦/20000 = log (5^2/4^2 ) log 𝑦/20000 = log 25/16 …(3) Removing log 𝑦/20000 = 25/16 y = 25/16×20000 y = 31250 ∴ Population of village in 2009 is 31250