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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 9.3, 12 Find a particular solution satisfying the given condition : 𝑥(𝑥^2−1) 𝑑𝑦/𝑑𝑥=1;𝑦=0 When 𝑥=2 𝑥(𝑥^2−1) dy = dx dy = 𝒅𝒙/(𝒙(𝒙𝟐 − 𝟏)) Integrating both sides. ∫1▒𝑑𝑦 = ∫1▒𝑑𝑥/(𝑥(𝑥2 − 1)) 𝒚 = ∫1▒𝒅𝒙/(𝒙(𝒙 + 𝟏)(𝒙 − 𝟏)) We can write integrand as 𝟏/(𝒙(𝒙 + 𝟏)(𝒙 − 𝟏)) = 𝑨/𝒙 + 𝒃/(𝒙 + 𝟏) + 𝒄/(𝒙 − 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/(𝒙(𝒙 + 𝟏)(𝒙 − 𝟏)) = (−𝟏)/𝒙 + 𝟏/(𝟐(𝒙 + 𝟏)) + 𝟏/(𝟐(𝒙 − 𝟏)) Now, From (1) y = ∫1▒1/(𝑥(𝑥 + 1)(𝑥 − 1)) dx = − ∫1▒𝟏/𝒙 + dx + 𝟏/𝟐 ∫1▒𝒅𝒙/(𝒙 + 𝟏) + 𝟏/𝟐 ∫1▒𝒅𝒙/(𝒙 − 𝟏) = log |𝒙|+ 𝟏/𝟐 log |𝒙+𝟏| + 𝟏/𝟐 log |𝒙−𝟏|+𝒄 = (−2)/2 log⁡|𝑥| + 𝟏/𝟐 log |𝒙+𝟏|+ 𝟏/𝟐 log |𝒙−𝟏|+𝑐 = 1/2 [−2 log⁡〖|𝑥|−2+𝐥𝐨𝐠⁡|(𝒙+𝟏)(𝒙−𝟏)| 〗 ]+𝑐 = 1/2 [log⁡〖𝑥^(−2)+log⁡|(𝑥+1)(𝑥−1)| 〗 ]+𝑐 = 1/2 [log⁡|𝑥^(−2) (𝑥^2−1)| ]+𝑐 = 𝟏/𝟐 log |(𝒙^𝟐 − 𝟏)/𝒙^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 = 𝟏/𝟐 log |(𝒙^𝟐 − 𝟏)/𝒙^𝟐 | − 𝟏/𝟐 log 𝟑/𝟒

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Davneet Singh

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.