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

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Misc 23 If ๐‘ฆ=๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ) , โ€“ 1 โ‰ค ๐‘ฅ โ‰ค 1, show that (1โˆ’๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 โˆ’๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ’ ๐‘Ž2 ๐‘ฆ =0 . ๐‘ฆ=๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ) Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ) )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ) ร— ๐‘‘(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ) ร— ๐‘Ž ((โˆ’1)/โˆš(1 โˆ’ ๐‘ฅ^2 )) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’๐‘Ž ๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ))/โˆš(1 โˆ’ ๐‘ฅ^2 ) โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’๐‘Ž๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ) โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’๐‘Ž๐‘ฆ Since we need to prove (1โˆ’๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 โˆ’ ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ’๐‘Ž2 ๐‘ฆ =0 Squaring (1) both sides (โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^2 = (โˆ’๐‘Ž๐‘ฆ)^2 (1โˆ’๐‘ฅ^2 ) (๐‘ฆ^โ€ฒ )^2 = ๐‘Ž^2 ๐‘ฆ^2 Differentiating again w.r.t x ๐‘‘((1 โˆ’ ๐‘ฅ^2 ) (๐‘ฆ^โ€ฒ )^2 )/๐‘‘๐‘ฅ = (d(๐‘Ž^2 ๐‘ฆ^2))/๐‘‘๐‘ฅ ๐‘‘((1 โˆ’ ๐‘ฅ^2 ) (๐‘ฆ^โ€ฒ )^2 )/๐‘‘๐‘ฅ = ๐‘Ž^2 (๐‘‘(๐‘ฆ^2))/๐‘‘๐‘ฅ ๐‘‘((1 โˆ’ ๐‘ฅ^2 ) (๐‘ฆ^โ€ฒ )^2 )/๐‘‘๐‘ฅ = ๐‘Ž^2 ร— 2๐‘ฆ ร—๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘(1 โˆ’ ๐‘ฅ^2 )/๐‘‘๐‘ฅ (๐‘ฆ^โ€ฒ )^2+(1 โˆ’ ๐‘ฅ^2 ) ๐’…((๐’š^โ€ฒ )^๐Ÿ )/๐’…๐’™ = ๐‘Ž^2 ร— 2๐‘ฆ๐‘ฆ^โ€ฒ (โˆ’2๐‘ฅ)(๐‘ฆ^โ€ฒ )^2+(1 โˆ’ ๐‘ฅ^2 )(๐Ÿ๐’š^โ€ฒ ร— ๐’…(๐’š^โ€ฒ )/๐’…๐’™) = ๐‘Ž^2 ร— 2๐‘ฆ๐‘ฆ^โ€ฒ (โˆ’2๐‘ฅ)(๐‘ฆ^โ€ฒ )^2+(1 โˆ’ ๐‘ฅ^2 )(๐Ÿ๐’š^โ€ฒ ร— ๐’š^โ€ฒโ€ฒ ) = ๐‘Ž^2 ร— 2๐‘ฆ๐‘ฆ^โ€ฒ Dividing both sides by ๐Ÿ๐’š^โ€ฒ โˆ’๐‘ฅ๐‘ฆ^โ€ฒ+(1 โˆ’ ๐‘ฅ^2 ) ๐‘ฆ^โ€ฒโ€ฒ = ๐‘Ž^2 ร— ๐‘ฆ โˆ’๐‘ฅ๐‘ฆ^โ€ฒ+(1 โˆ’ ๐‘ฅ^2 ) ๐‘ฆ^โ€ฒโ€ฒ = ๐‘Ž^2 ๐‘ฆ (๐Ÿ โˆ’ ๐’™^๐Ÿ ) ๐’š^โ€ฒโ€ฒโˆ’๐’™๐’š^โ€ฒโˆ’๐’‚^๐Ÿ ๐’š=๐ŸŽ Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.