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Ex 9.3
Ex 9.3, 2
Ex 9.3, 3
Ex 9.3, 4 Important
Ex 9.3, 5
Ex 9.3, 6
Ex 9.3, 7 Important
Ex 9.3, 8
Ex 9.3, 9 Important
Ex 9.3, 10 Important
Ex 9.3, 11 Important
Ex 9.3, 12 You are here
Ex 9.3, 13
Ex 9.3, 14
Ex 9.3, 15 Important
Ex 9.3, 16
Ex 9.3, 17 Important
Ex 9.3, 18
Ex 9.3, 19 Important
Ex 9.3, 20 Important
Ex 9.3, 21
Ex 9.3, 22 Important
Ex 9.3, 23 (MCQ)
Last updated at Aug. 11, 2023 by Teachoo
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