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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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.