# Ex 9.4, 12

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.4, 12 Find a particular solution satisfying the given condition : 𝑥 𝑥2−1 𝑑𝑦𝑑𝑥=1;𝑦=0 When 𝑥=2 𝑥 𝑥2−1 dy = dx dy = 𝑑𝑥𝑥(𝑥2 − 1) Integrating both sides. 𝑑𝑦 = 𝑑𝑥𝑥(𝑥2 − 1) 𝑦 = 𝑑𝑥𝑥(𝑥 + 1)(𝑥 − 1) We can write integrand as 1𝑥(𝑥 + 1)(𝑥 − 1) = 𝐴𝑥 + 𝑏𝑥 + 1 + 𝑐𝑥 − 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 = 12 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 = 12 Therefore 1𝑥(𝑥 + 1)(𝑥 − 1) = −1𝑥 + 12(𝑥 + 1) + 12(𝑥 − 1) Now, From (1) y = 1𝑥(𝑥 + 1)(𝑥 − 1) dx = − 1𝑥 + dx + 12 𝑑𝑥𝑥 + 1 + 12 𝑑𝑥𝑥 − 1 = log 𝑥+ 12 log 𝑥+1 + 12 log 𝑥−1+𝑐 = −22 log 𝑥 + 12 log 𝑥+1+ 12 log 𝑥−1+𝑐 = 12 −2 log 𝑥−2+ log (𝑥+1)(𝑥−1)+𝑐 = 12 log 𝑥−2+ log (𝑥+1)(𝑥−1)+𝑐 = 12 log 𝑥−2( 𝑥2−1)+𝑐 = 12 log 𝑥2 − 1 𝑥2+𝑐 ⇒ y = 12 log 𝑥2 − 1 𝑥2 + C Given that x = 2, y = 0 Substituting values in (1) we get 0 = 12 log 22−1 22 + C 0 = 12 log 34 + C C = − 12 log 34 Putting value of c in (1), y = 𝟏𝟐 log 𝒙𝟐 − 𝟏 𝒙𝟐 − 𝟏𝟐 log 𝟑𝟒

Chapter 9 Class 12 Differential Equations

Serial order wise

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