# Example 14 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by Teachoo

Last updated at Dec. 11, 2019 by Teachoo

Transcript

Example 14 In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself ? Rate of change of principal = 5% of Principal ๐๐/๐๐ก = 5% ร P ๐๐/๐๐ก = 5/100 ร P ๐๐/๐๐ก = 1/20 ร P ๐๐/๐ = ๐๐ก/20 Integrating both sides โซ1โ๐๐/๐ = โซ1โ๐๐ก/20 log |๐| = ๐ก/20 + C Removing log P = e^(๐ก/20 + ๐ถ) ร ec P = e^(๐ก/20 + ๐ถ) ร ec P = ke^(๐ก/20) where k = ec Now, we have to find in how many years Rs 1000 double it self Thus, we need to find time T when Principal is Rs 2000 First let us find k At t = 0, P = 1000 Putting in (1) P = ke^(๐ก/20) 1000 = ke^(0/20) 1000 = k e0 1000 = k ร 1 1000 = k So, k = 1000 Put k = 1000 in (1) P = ke^(๐ก/20) P = 1000 e^(๐ก/20) Now, we need to find time t when Principal is Rs 2000 Putting P = 2000, t = t 2000 = 1000 e^(๐ก/20) 2000/1000 = e^(๐ก/20) 2 = e^(๐ก/20) e^(๐ก/20) = 2 Taking log both sides log_๐โกใ๐^(๐ก/20) ใ=log_๐โก2 t/20 log_๐โก๐=log_๐โก2 " " t/20ร1=log_๐โก2 ๐ญ=๐๐ ใ๐๐๐ใ_๐โก๐ (log x๐ = ๐log x) (log e = 1)

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Chapter 9 Class 12 Differential Equations (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.