# Ex 9.4, 12 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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

Ex 9.4

Ex 9.4, 1
Important

Ex 9.4, 2

Ex 9.4, 3

Ex 9.4, 4 Important

Ex 9.4, 5

Ex 9.4, 6

Ex 9.4, 7

Ex 9.4, 8

Ex 9.4, 9 Important

Ex 9.4, 10 Important

Ex 9.4, 11

Ex 9.4, 12 You are here

Ex 9.4, 13

Ex 9.4, 14

Ex 9.4, 15 Important

Ex 9.4, 16

Ex 9.4, 17

Ex 9.4, 18

Ex 9.4, 19 Important

Ex 9.4, 20 Important

Ex 9.4, 21

Ex 9.4, 22 Important

Ex 9.4, 23

Chapter 9 Class 12 Differential Equations

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.