Ex 9.3

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.3, 12 Find a particular solution satisfying the given condition : ๐ฅ(๐ฅ^2โ1) ๐๐ฆ/๐๐ฅ=1;๐ฆ=0 When ๐ฅ=2 ๐ฅ(๐ฅ^2โ1) dy = dx dy = ๐๐/(๐(๐๐ โ ๐)) Integrating both sides. โซ1โ๐๐ฆ = โซ1โ๐๐ฅ/(๐ฅ(๐ฅ2 โ 1)) ๐ = โซ1โ๐๐/(๐(๐ + ๐)(๐ โ ๐)) We can write integrand as ๐/(๐(๐ + ๐)(๐ โ ๐)) = ๐จ/๐ + ๐/(๐ + ๐) + ๐/(๐ โ ๐) By canceling the denominators. 1 = A (x โ 1) (x + 1) B x (x โ 1) + C x (x + 1) Putting x = 0 1 = A (0 โ 1) (0 + 1) + B.0. (0 โ 1) + C.0. (0 + 1) 1 = A (โ1) (1) + B.0 + C.0 1 = โ A A = โ1 Similarly putting x = โ1 1 = A (โ1 โ 1) (โ1 + 1) + B (โ1) (โ1 โ 1) + C(โ1)(โ1 + 1) 1 = A (โ2) (0) + B (โ1) (โ2) + C (โ1) (0) 1 = 0 + 2B + 0 2B = 1 B = ๐/๐ Similarly putting x = 1 1 = A(1 โ 1) (1 + 1) + B.1(1 โ 1) + C(1)(1 + 1) 1 = A (0) (2) + B.1.0 + C.2 2C = 1 C = ๐/๐ Therefore, ๐/(๐(๐ + ๐)(๐ โ ๐)) = (โ๐)/๐ + ๐/(๐(๐ + ๐)) + ๐/(๐(๐ โ ๐)) Now, From (1) y = โซ1โ1/(๐ฅ(๐ฅ + 1)(๐ฅ โ 1)) dx = โ โซ1โ๐/๐ + dx + ๐/๐ โซ1โ๐๐/(๐ + ๐) + ๐/๐ โซ1โ๐๐/(๐ โ ๐) = log |๐|+ ๐/๐ log |๐+๐| + ๐/๐ log |๐โ๐|+๐ = (โ2)/2 logโก|๐ฅ| + ๐/๐ log |๐+๐|+ ๐/๐ log |๐โ๐|+๐ = 1/2 [โ2 logโกใ|๐ฅ|โ2+๐ฅ๐จ๐ โก|(๐+๐)(๐โ๐)| ใ ]+๐ = 1/2 [logโกใ๐ฅ^(โ2)+logโก|(๐ฅ+1)(๐ฅโ1)| ใ ]+๐ = 1/2 [logโก|๐ฅ^(โ2) (๐ฅ^2โ1)| ]+๐ = ๐/๐ log |(๐^๐ โ ๐)/๐^๐ |+๐ Given that x = 2, y = 0 Substituting values in (1) we get 0 = 1/2 " log " |(2^2โ1)/2^2 |" + C" 0 = 1/2 " log " 3/4 " + C" C = โ๐/๐ " log " ๐/๐ Putting value of c in (1), y = ๐/๐ log |(๐^๐ โ ๐)/๐^๐ | โ ๐/๐ log ๐/๐