1. Chapter 9 Class 12 Differential Equations
2. Serial order wise

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Ex 9.4, 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. ﷮﷮𝑑𝑦﷯ = ﷮﷮ 𝑑𝑥﷮𝑥(𝑥2 − 1)﷯﷯ 𝑦 = ﷮﷮ 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﷮𝑥(𝑥 + 1)(𝑥 − 1)﷯﷯ dx = − ﷮﷮ 1﷮𝑥﷯﷯ + dx + 1﷮2﷯ ﷮﷮ 𝑑𝑥﷮𝑥 + 1﷯﷯ + 1﷮2﷯ ﷮﷮ 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 ⇒ y = 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﷯