Ex 9.3

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.3, 20 In a bank, principal increases continuously at the rate of ๐% per year. Find the value of r if Rs 100 double itself in 10 years (logโกใ2=0.6931ใ )Let Principal = p Given, principal increases ar rate r % per year โด ๐๐/๐๐ = ๐ % ร P โด ๐๐/๐๐ก = ๐/100 ร p ๐๐/๐ = ๐/๐๐๐ dt Integrating both sides โซ1โ๐๐/๐ = ๐/100 โซ1โ๐๐ก log p = ๐๐/๐๐๐ + log c log p โ log c = ๐๐ก/100 log ๐/๐ = ๐๐ก/100 ๐/๐ = ๐^(๐๐/๐๐๐) As we have put Rs 100 initially Putting t = 0 and p = 100 in (1) 100/๐ = ๐^((๐ ร 0)/100) 100/๐ = e^0 100/๐ = 1 c = 100 Putting value of C in equation (1) ๐/(๐๐๐ ) = ๐^(๐๐/๐๐๐) Also, given that Rs 100 will double itself in 10 years โด Putting t = 10, p = 200 in the equation 200/(100 ) = e^(10๐/100) 2 = e^(๐/10) log 2 = ๐/10 0.6931 = ๐/10 r = 6.931 โด Rate of interest = r = 6.931 %