Ex 9.4, 12 - Chapter 9 Class 12 Differential Equations (Term 2)

Transcript

Ex 9.4, 12 Find a particular solution satisfying the given condition : 2 1 =1; =0 When =2 2 1 dy = dx dy = ( 2 1) Integrating both sides. = ( 2 1) = ( + 1)( 1) We can write integrand as 1 ( + 1)( 1) = + + 1 + 1 By canceling the denominators. 1 = A (x 1) (x + 1) B x (x 1) + C x (x + 1) Putting x = 0 1 = A (0 1) (0 + 1) + B.0. (0 1) + C.0. (0 + 1) 1 = A ( 1) (1) + B.0 + C.0 1 = A A = 1 Similarly putting x = 1 1 = A ( 1 1) ( 1 + 1) + B ( 1) ( 1 1) + C( 1)( 1 + 1) 1 = A ( 2) (0) + B ( 1) ( 2) + C ( 1) (0) 1 = 0 + 2B + 0 2B = 1 B = 1 2 Similarly putting x = 1 1 = A(1 1) (1 + 1) + B.1(1 1) + C(1)(1 + 1) 1 = A (0) (2) + B.1.0 + C.2 2C = 1 C = 1 2 Therefore 1 ( + 1)( 1) = 1 + 1 2( + 1) + 1 2( 1) Now, From (1) y = 1 ( + 1)( 1) dx = 1 + dx + 1 2 + 1 + 1 2 1 = log + 1 2 log +1 + 1 2 log 1 + = 2 2 log + 1 2 log +1 + 1 2 log 1 + = 1 2 2 log 2+ log ( +1)( 1) + = 1 2 log 2 + log ( +1)( 1) + = 1 2 log 2 ( 2 1) + = 1 2 log 2 1 2 + y = 1 2 log 2 1 2 + C Given that x = 2, y = 0 Substituting values in (1) we get 0 = 1 2 log 2 2 1 2 2 + C 0 = 1 2 log 3 4 + C C = 1 2 log 3 4 Putting value of c in (1), y = log log