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Misc 23 - If y = ea cos-1 x, show (1 - x2) d2y/dx2 - x dy/dx

Misc 23 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Misc 23 If 𝑦=𝑒^(γ€–π‘Ž π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯) , – 1 ≀ π‘₯ ≀ 1, show that (1βˆ’π‘₯^2 ) (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 βˆ’π‘₯ 𝑑𝑦/𝑑π‘₯ βˆ’ π‘Ž2 𝑦 =0 . 𝑦=𝑒^(γ€–π‘Ž π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = 𝑑(𝑒^(γ€–π‘Ž π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯" " ) )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑒^(γ€–π‘Ž π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯" " ) Γ— 𝑑(γ€–π‘Ž π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑒^(γ€–π‘Ž π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯" " ) Γ— π‘Ž ((βˆ’1)/√(1 βˆ’ π‘₯^2 )) 𝑑𝑦/𝑑π‘₯ = (βˆ’π‘Ž 𝑒^(γ€–π‘Ž π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯" " ))/√(1 βˆ’ π‘₯^2 ) √(1 βˆ’ π‘₯^2 ) 𝑑𝑦/𝑑π‘₯ = βˆ’π‘Žπ‘’^(γ€–π‘Ž π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯" " ) √(1 βˆ’ π‘₯^2 ) 𝑑𝑦/𝑑π‘₯ = βˆ’π‘Žπ‘¦ Since we need to prove (1βˆ’π‘₯^2 ) (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 βˆ’ π‘₯ 𝑑𝑦/𝑑π‘₯ βˆ’π‘Ž2 𝑦 =0 Squaring (1) both sides (√(1 βˆ’ π‘₯^2 ) 𝑑𝑦/𝑑π‘₯)^2 = (βˆ’π‘Žπ‘¦)^2 (1βˆ’π‘₯^2 ) (𝑦^β€² )^2 = π‘Ž^2 𝑦^2 Differentiating again w.r.t x 𝑑((1 βˆ’ π‘₯^2 ) (𝑦^β€² )^2 )/𝑑π‘₯ = (d(π‘Ž^2 𝑦^2))/𝑑π‘₯ 𝑑((1 βˆ’ π‘₯^2 ) (𝑦^β€² )^2 )/𝑑π‘₯ = π‘Ž^2 (𝑑(𝑦^2))/𝑑π‘₯ 𝑑((1 βˆ’ π‘₯^2 ) (𝑦^β€² )^2 )/𝑑π‘₯ = π‘Ž^2 Γ— 2𝑦 ×𝑑𝑦/𝑑π‘₯ 𝑑(1 βˆ’ π‘₯^2 )/𝑑π‘₯ (𝑦^β€² )^2+(1 βˆ’ π‘₯^2 ) 𝒅((π’š^β€² )^𝟐 )/𝒅𝒙 = π‘Ž^2 Γ— 2𝑦𝑦^β€² (βˆ’2π‘₯)(𝑦^β€² )^2+(1 βˆ’ π‘₯^2 )(πŸπ’š^β€² Γ— 𝒅(π’š^β€² )/𝒅𝒙) = π‘Ž^2 Γ— 2𝑦𝑦^β€² (βˆ’2π‘₯)(𝑦^β€² )^2+(1 βˆ’ π‘₯^2 )(πŸπ’š^β€² Γ— π’š^β€²β€² ) = π‘Ž^2 Γ— 2𝑦𝑦^β€² Dividing both sides by πŸπ’š^β€² βˆ’π‘₯𝑦^β€²+(1 βˆ’ π‘₯^2 ) 𝑦^β€²β€² = π‘Ž^2 Γ— 𝑦 βˆ’π‘₯𝑦^β€²+(1 βˆ’ π‘₯^2 ) 𝑦^β€²β€² = π‘Ž^2 𝑦 (𝟏 βˆ’ 𝒙^𝟐 ) π’š^β€²β€²βˆ’π’™π’š^β€²βˆ’π’‚^𝟐 π’š=𝟎 Hence proved

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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.