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Ex 9.5, 19 The integrating Factor of the differential equation \(1−𝑦^2 ) 𝑑𝑥/𝑑𝑦+𝑦𝑥=𝑎𝑦 (−1<𝑦<1) is (A) 1/(𝑦^2−1) (B) 1/√(𝑦^2−1) (C) 1/(1−𝑦^2 ) (D) 1/√(1−𝑦^2 ) (1−𝑦^2 ) 𝑑𝑥/𝑑𝑦+𝑦𝑥=𝑎𝑦 Dividing both sides by 1 − y2 𝑑𝑥/𝑑𝑦 + 𝑦𝑥/(1−𝑦^2 ) = 𝑎𝑦/(1−𝑦^2 ) Differential equation is of the form 𝒅𝒙/𝒅𝒚 + P1x = Q1 where P1 = 𝒚/(𝟏 − 𝒚^𝟐 ) & Q1 = 𝒂𝒚/(𝟏 − 𝒚^𝟐 ) IF = 𝒆^∫1▒𝒑𝟏𝒅𝒚 Finding ∫1▒〖𝑷𝟏 𝒅𝒚〗 ∫1▒〖𝑃1 𝑑𝑦=〗 ∫1▒〖𝑦/(1−𝑦^2 ) 𝑑𝑦 〗 Putting 1 − y2 = t −2y dy = dt y dy = (−1)/2 dt ∴ Our equation becomes ∫1▒〖𝑃1 𝑑𝑦= (−1)/2 〗 ∫1▒〖𝑑𝑡/𝑡 〗 ∫1▒〖𝑃1 𝑑𝑦= (−1)/2 〗 log⁡𝑡 Putting back value of t ∫1▒〖𝑃1 𝑑𝑦= (−1)/2 〗 log⁡(1−𝑦^2) ∫1▒〖𝑃1 𝑑𝑦=〗 〖log⁡(1−𝑦^2 )〗^((−1)/2) ∫1▒〖𝑷𝟏 𝒅𝒚= 〗 𝐥𝐨𝐠 𝟏/√(𝟏 − 𝒚^𝟐 ) Thus, IF = 𝑒^∫1▒𝑝1𝑑𝑥 IF = 𝑒^(𝑙𝑜𝑔 1/√(1 − 𝑦^2 )) IF = 𝟏/√(𝟏 − 𝒚^𝟐 ) So, the correct answer is (d)

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo