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Misc 23 If 𝑦=𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥) , – 1 ≤ 𝑥 ≤ 1, show that (1−𝑥^2 ) (𝑑^2 𝑦)/〖𝑑𝑥〗^2 −𝑥 𝑑𝑦/𝑑𝑥 − 𝑎2 𝑦 =0 . 𝑦=𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥) Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑥 = 𝑑(𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ) )/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ) × 𝑑(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥)/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ) × 𝑎 ((−1)/√(1 − 𝑥^2 )) 𝑑𝑦/𝑑𝑥 = (−𝑎 𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ))/√(1 − 𝑥^2 ) √(1 − 𝑥^2 ) 𝑑𝑦/𝑑𝑥 = −𝑎𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ) √(1 − 𝑥^2 ) 𝑑𝑦/𝑑𝑥 = −𝑎𝑦 Since we need to prove (1−𝑥^2 ) (𝑑^2 𝑦)/〖𝑑𝑥〗^2 − 𝑥 𝑑𝑦/𝑑𝑥 −𝑎2 𝑦 =0 Squaring (1) both sides (√(1 − 𝑥^2 ) 𝑑𝑦/𝑑𝑥)^2 = (−𝑎𝑦)^2 (1−𝑥^2 ) (𝑦^′ )^2 = 𝑎^2 𝑦^2 Differentiating again w.r.t x 𝑑((1 − 𝑥^2 ) (𝑦^′ )^2 )/𝑑𝑥 = (d(𝑎^2 𝑦^2))/𝑑𝑥 𝑑((1 − 𝑥^2 ) (𝑦^′ )^2 )/𝑑𝑥 = 𝑎^2 (𝑑(𝑦^2))/𝑑𝑥 𝑑((1 − 𝑥^2 ) (𝑦^′ )^2 )/𝑑𝑥 = 𝑎^2 × 2𝑦 ×𝑑𝑦/𝑑𝑥 𝑑(1 − 𝑥^2 )/𝑑𝑥 (𝑦^′ )^2+(1 − 𝑥^2 ) 𝒅((𝒚^′ )^𝟐 )/𝒅𝒙 = 𝑎^2 × 2𝑦𝑦^′ (−2𝑥)(𝑦^′ )^2+(1 − 𝑥^2 )(𝟐𝒚^′ × 𝒅(𝒚^′ )/𝒅𝒙) = 𝑎^2 × 2𝑦𝑦^′ (−2𝑥)(𝑦^′ )^2+(1 − 𝑥^2 )(𝟐𝒚^′ × 𝒚^′′ ) = 𝑎^2 × 2𝑦𝑦^′ Dividing both sides by 𝟐𝒚^′ −𝑥𝑦^′+(1 − 𝑥^2 ) 𝑦^′′ = 𝑎^2 × 𝑦 −𝑥𝑦^′+(1 − 𝑥^2 ) 𝑦^′′ = 𝑎^2 𝑦 (𝟏 − 𝒙^𝟐 ) 𝒚^′′−𝒙𝒚^′−𝒂^𝟐 𝒚=𝟎 Hence proved

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.