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Last updated at May 29, 2018 by Teachoo
Transcript
Ex 9.6, 19 The integrating Factor of the differential equation 1− 𝑦2 𝑑𝑥𝑑𝑦+𝑦𝑥=𝑎𝑦 −1<𝑦<1 is (A) 1 𝑦2−1 (B) 1 𝑦2−1 (C) 11− 𝑦2 (D) 1 1− 𝑦2 1− 𝑦2 𝑑𝑦𝑑𝑥+𝑦𝑥=𝑎𝑦 Dividing both sides by 1 − y2 𝑑𝑦𝑑𝑥 + 𝑦𝑥1− 𝑦2 = 𝑎𝑦1− 𝑦2 Differential equation is of the form 𝑑𝑦𝑑𝑥 + P1x = Q1 where P1 = 𝑦1 − 𝑦2 & Q1 = 𝑎𝑦1 − 𝑦2 IF = 𝑒 𝑝1𝑑𝑥 Finding 𝑷𝟏 𝒅𝒚 𝑃1 𝑑𝑦= 𝑦1− 𝑦2 𝑑𝑦 Putting 1 − y2 = t −2y dy = dt y dy = −12 dt ∴ Our equation becomes 𝑃1 𝑑𝑦= −12 𝑑𝑡𝑡 𝑃1 𝑑𝑦= −12 log𝑡 Putting back value of t 𝑃1 𝑑𝑦= −12 log(1− 𝑦2) 𝑃1 𝑑𝑦= log 1− 𝑦2 −12 𝑃1 𝑑𝑦= log 1 1 − 𝑦2 Thus, IF = 𝑒 𝑝1𝑑𝑥 IF = elog 1 1 − 𝑦2 IF = 1 1 − 𝑦2 ∴ Part (D) is correct answer.
Ex 9.6
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