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Ex 9.4, 12  Find particular solution: x (x2 - 1) dy/dx = 1, y=0 - Ex 9.4

Ex 9.4, 12 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 12 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.4, 12 - Chapter 9 Class 12 Differential Equations - Part 4 Ex 9.4, 12 - Chapter 9 Class 12 Differential Equations - Part 5

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.