Question 17

If y = cos (m cos -1 ⁑x)

Show that : (1 - x 2 ) d 2 y/dx 2 – x dy/dx + m 2 y = 0

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Question 17 If y = cos⁑〖(π‘š cos^(βˆ’1)⁑π‘₯)γ€— Show that : (1βˆ’π‘₯^2) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) – π‘₯ 𝑑𝑦/𝑑π‘₯ + π‘š^2 𝑦 = 0 y = cos⁑〖(π‘š cos^(βˆ’1)⁑π‘₯)γ€— Differentiating w.r.t. x 𝑑𝑦/𝑑π‘₯ = 𝑑(cos⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) 𝑑(π‘š cos^(βˆ’1)⁑π‘₯ )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—π‘š 𝑑(cos^(βˆ’1)⁑π‘₯ )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—π‘šΓ—(βˆ’1)/√(1 βˆ’ π‘₯^2 ) 𝑑𝑦/𝑑π‘₯ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—1/√(1 βˆ’ π‘₯^2 ) 𝑦′ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—1/√(1 βˆ’ π‘₯^2 ) Now, finding 𝑦′′ will be complicated So, we multiply √(𝟏 βˆ’ 𝒙^𝟐 ) to the left side √(1βˆ’π‘₯^2 ) 𝑦′ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) Squaring both sides (√(1βˆ’π‘₯^2 ) )^2 (𝑦^β€² )^2 = (βˆ’m)^2 sin^2⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) 𝑑𝑦/𝑑π‘₯ = sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—π‘šΓ—(βˆ’1)/√(1 βˆ’ π‘₯^2 ) 𝑑𝑦/𝑑π‘₯ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—1/√(1 βˆ’ π‘₯^2 ) 𝑦′ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—1/√(1 βˆ’ π‘₯^2 ) Now, finding 𝑦′′ will be complicated So, we multiply √(𝟏 βˆ’ 𝒙^𝟐 ) to the left side √(1βˆ’π‘₯^2 ) 𝑦′ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) Squaring both sides (√(1βˆ’π‘₯^2 ) )^2 (𝑦^β€² )^2 = (βˆ’m)^2 sin^2⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) 𝑑𝑦/𝑑π‘₯ = sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—π‘šΓ—(βˆ’1)/√(1 βˆ’ π‘₯^2 ) 𝑑𝑦/𝑑π‘₯ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—1/√(1 βˆ’ π‘₯^2 ) 𝑦′ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ )Γ—1/√(1 βˆ’ π‘₯^2 ) Now, finding 𝑦′′ will be complicated So, we multiply √(𝟏 βˆ’ 𝒙^𝟐 ) to the left side √(1βˆ’π‘₯^2 ) 𝑦′ = βˆ’m sin⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) Squaring both sides (√(1βˆ’π‘₯^2 ) )^2 (𝑦^β€² )^2 = (βˆ’m)^2 sin^2⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) (1βˆ’x^2 ) (𝑦^β€² )^2 = π‘š^2 sin^2⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) We know that sin2 x = 1 – cos2 x (1βˆ’x^2 ) (𝑦^β€² )^2 = π‘š^2 (1βˆ’cos^2⁑(π‘š cos^(βˆ’1)⁑π‘₯ ) ) Putting y = cos⁑〖(π‘š cos^(βˆ’1)⁑π‘₯)γ€— (1βˆ’x^2 ) (𝑦^β€² )^2 = π‘š^2 (1βˆ’π‘¦^2 ) Differentiating again w.r.t. x 𝑑(1 βˆ’ x^2 )/𝑑π‘₯ (𝑦^β€² )^2 + (1βˆ’x^2 ) (𝑑(𝑦^β€² )^2)/𝑑π‘₯ = π‘š^2 𝑑(1 βˆ’ 𝑦^2 )/𝑑π‘₯ –2x (𝑦^β€² )^2 + (1βˆ’x^2 )2𝑦^β€² 𝑦′′ = π‘š^2Γ—βˆ’2𝑦𝑦′ 2𝑦^β€² (βˆ’π‘₯𝑦^β€² "+ " (1βˆ’x^2 )𝑦′′) = βˆ’2y^β€²Γ—π‘š^2 𝑦 βˆ’π‘₯𝑦^β€² "+ " (1βˆ’x^2 )𝑦′′ = βˆ’π‘š^2 𝑦 (1βˆ’x^2 )𝑦′′ βˆ’ π‘₯𝑦^β€² + π‘š^2 𝑦 = 0 (1βˆ’π‘₯^2) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) – π‘₯ 𝑑𝑦/𝑑π‘₯ + π‘š^2 𝑦 = 0 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.