Examples

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Example 9 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 ๐๐ท/๐๐ = 5% ร P ๐๐/๐๐ก = 5/100 ร P ๐๐/๐๐ก = 1/20 ร P ๐๐/๐ท = ๐๐/๐๐ Integrating both sides โซ1โ๐๐/๐ = โซ1โ๐๐ก/20 log |๐ท| = ๐/๐๐ + C Removing log P = e^(๐ก/20 + ๐ถ) ร ec P = e^(๐ก/20 + ๐ถ) ร ec P = k๐^(๐/๐๐) 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^(๐ก/20) 1000 = k๐^(๐/๐๐) 1000 = k e0 1000 = k ร 1 1000 = k So, k = 1000 Put k = 1000 in (1) P = ke^(๐ก/20) P = 1000 ๐^(๐/๐๐) Now, we need to find time t when Principal is Rs 2000 Putting P = 2000, t = t 2000 = 1000 e^(๐ก/20) 2000/1000 = e^(๐ก/20) 2 = e^(๐ก/20) ๐^(๐/๐๐) = 2 Taking log both sides ใ๐ฅ๐จ๐ ใ_๐โกใ๐^(๐/๐๐) ใ=ใ๐ฅ๐จ๐ ใ_๐โก๐ t/20 log_๐โก๐=log_๐โก2 " " t/20ร 1=log_๐โก2 ๐ญ=๐๐ ใ๐๐๐ใ_๐โก๐