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Ex 9.6, 19 - Class 12 - Part 1 - Find Integrating Factor.jpeg

Ex 9.6, 19 - Class 12 - Part 2 - Integrating w.r.t y.jpeg
Ex 9.6, 19 - Class 12 - Part 3 - The correct answer is.jpeg

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Ex 9.6, 19 The integrating Factor of the differential equation (1βˆ’π‘¦^2 ) 𝑑π‘₯/𝑑𝑦+𝑦π‘₯=π‘Žπ‘¦ (βˆ’1<𝑦<1) is (A) 1/(𝑦^2βˆ’1) (B) 1/√(𝑦^2βˆ’1) (C) 1/(1βˆ’π‘¦^2 ) (D) 1/√(1βˆ’π‘¦^2 ) (1βˆ’π‘¦^2 ) 𝑑π‘₯/𝑑𝑦+𝑦π‘₯=π‘Žπ‘¦ Dividing both sides by 1 βˆ’ y2 𝑑π‘₯/𝑑𝑦 + 𝑦π‘₯/(1βˆ’π‘¦^2 ) = π‘Žπ‘¦/(1βˆ’π‘¦^2 ) Differential equation is of the form 𝒅𝒙/π’…π’š + P1x = Q1 where P1 = 𝑦/(1 βˆ’ 𝑦^2 ) & Q1 = π‘Žπ‘¦/(1 βˆ’ 𝑦^2 ) IF = 𝑒^∫1▒𝑝1𝑑𝑦 Finding ∫1β–’γ€–π‘·πŸ π’…π’šγ€— ∫1▒〖𝑃1 𝑑𝑦=γ€— ∫1▒〖𝑦/(1βˆ’π‘¦^2 ) 𝑑𝑦 γ€— Putting 1 βˆ’ y2 = t βˆ’2y dy = dt y dy = (βˆ’1)/2 dt ∴ Our equation becomes ∫1▒〖𝑃1 𝑑𝑦= (βˆ’1)/2 γ€— ∫1▒〖𝑑𝑑/𝑑 γ€— ∫1▒〖𝑃1 𝑑𝑦= (βˆ’1)/2 γ€— log⁑𝑑 Putting back value of t ∫1▒〖𝑃1 𝑑𝑦= (βˆ’1)/2 γ€— log⁑(1βˆ’π‘¦^2) ∫1▒〖𝑃1 𝑑𝑦=γ€— γ€–log⁑(1βˆ’π‘¦^2 )γ€—^((βˆ’1)/2) ∫1▒〖𝑃1 𝑑𝑦= γ€— log 1/√(1 βˆ’ 𝑦^2 ) Thus, IF = 𝑒^∫1▒𝑝1𝑑π‘₯ IF = 𝑒^(π‘™π‘œπ‘” 1/√(1 βˆ’ 𝑦^2 )) IF = 𝟏/√(𝟏 βˆ’ π’š^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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.