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

If y = cos (m cos^-1 ⁡x)  Show that : (1 - x^2) y'' - x y'+ m^2 y = 0

Question 17 - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 17 - CBSE Class 12 Sample Paper for 2019 Boards - Part 3
Question 17 - CBSE Class 12 Sample Paper for 2019 Boards - Part 4

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Transcript

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 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 and Computer Science at Teachoo