Check sibling questions

Example 41 - If y = sin-1 x, show that (1 - x2) d2y/dx2 - x dy/dx = 0

Example 41 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Example 41 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

Example 41 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Example 41 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

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Example 41 (Method 1) If y = 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯, show that (1 – π‘₯2) 𝑑2𝑦/𝑑π‘₯2 βˆ’ π‘₯ 𝑑𝑦/𝑑π‘₯ = 0 . We have 𝑦 = 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/√(γ€–1 βˆ’ π‘₯γ€—^2 ) √((πŸβˆ’π’™^𝟐 ) ) π’š^β€² = 𝟏 Squaring both sides ("As " 𝑑(〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯)/𝑑π‘₯ " = " 1/√(γ€–1 βˆ’ π‘₯γ€—^2 )) (√((1βˆ’π‘₯^2 ) ) 𝑦^β€² )^2 = 1^2 (1βˆ’π‘₯^2 )(𝑦^β€² )^2 = 1 Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ ((1βˆ’π‘₯^2 )(𝑦^β€² )^2 ) = (𝑑(1))/𝑑π‘₯ d(1 βˆ’ x^2 )/𝑑π‘₯ (𝑦^β€² )^2+(1βˆ’π‘₯^2 ) 𝑑((𝑦^β€² )^2 )/𝑑π‘₯ = 0 βˆ’2π‘₯(𝑦^β€² )^2+(1βˆ’π‘₯^2 ) 2𝑦^β€² Γ— 𝑦^β€²β€² = 0 γ€–2yγ€—^β€² [βˆ’π’™π’š^β€²+(πŸβˆ’π’™^𝟐 ) π’š^β€²β€² ] = 0 βˆ’π‘₯𝑦^β€²+(1βˆ’π‘₯^2 ) 𝑦^β€²β€²=0 (γ€–πŸβˆ’π’™γ€—^𝟐 ) (𝒅^𝟐 π’š)/〖𝒅𝒙〗^𝟐 βˆ’ 𝒙 . π’…π’š/𝒅𝒙 = 0 (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 = (βˆ’1)/( 2) (γ€–1βˆ’π‘₯γ€—^2 )^((βˆ’3)/2 ). (0βˆ’2π‘₯) (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 = (βˆ’1)/( 2) (γ€–1βˆ’π‘₯γ€—^2 )^((βˆ’3)/2 ). (βˆ’2π‘₯) (𝒅^𝟐 π’š)/〖𝒅𝒙〗^𝟐 = 𝒙(γ€–πŸβˆ’π’™γ€—^𝟐 )^((βˆ’πŸ‘)/𝟐 ) Example 41 (Method 2) If y = 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯, show that (1 – π‘₯2) 𝑑2𝑦/𝑑π‘₯2 βˆ’ π‘₯ 𝑑𝑦/𝑑π‘₯ = 0 . We have 𝑦 = 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/√(γ€–1 βˆ’ π‘₯γ€—^2 ) π’…π’š/𝒅𝒙 = (γ€–πŸβˆ’π’™γ€—^𝟐 )^((βˆ’πŸ)/( 𝟐)) ("As " 𝑑(〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯)/𝑑π‘₯ " = " 1/√(γ€–1 βˆ’ π‘₯γ€—^2 )) Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = (𝑑(γ€–1 βˆ’ π‘₯γ€—^2 )^((βˆ’1)/( 2)))/𝑑π‘₯ (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 = (βˆ’1)/( 2) (γ€–1βˆ’π‘₯γ€—^2 )^((βˆ’1)/( 2) βˆ’1) . 𝑑(γ€–1 βˆ’ π‘₯γ€—^2 )/𝑑π‘₯ (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 = (βˆ’1)/( 2) (γ€–1βˆ’π‘₯γ€—^2 )^((βˆ’3)/2 ). (0βˆ’2π‘₯) (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 = (βˆ’1)/( 2) (γ€–1βˆ’π‘₯γ€—^2 )^((βˆ’3)/2 ). (βˆ’2π‘₯) (𝒅^𝟐 π’š)/〖𝒅𝒙〗^𝟐 = 𝒙(γ€–πŸβˆ’π’™γ€—^𝟐 )^((βˆ’πŸ‘)/𝟐 ) Now, We need to prove (γ€–1βˆ’π‘₯γ€—^2 ) (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 βˆ’ π‘₯ . 𝑑𝑦/𝑑π‘₯ = 0 Solving LHS (γ€–1βˆ’π‘₯γ€—^2 ) (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 βˆ’ π‘₯ . 𝑑𝑦/𝑑π‘₯ = (γ€–1βˆ’π‘₯γ€—^2 ) . (π‘₯γ€– (γ€–1βˆ’π‘₯γ€—^2 )γ€—^((βˆ’3)/2 ) ) βˆ’ π‘₯ (γ€–1βˆ’π‘₯γ€—^2 )^((βˆ’1)/( 2)) = π‘₯γ€– (γ€–1βˆ’π‘₯γ€—^2 )γ€—^(𝟏 + ((βˆ’πŸ‘)/𝟐) )βˆ’π‘₯ (γ€–1βˆ’π‘₯γ€—^2 )^((βˆ’1)/( 2)) = π‘₯γ€– (γ€–1βˆ’π‘₯γ€—^2 )γ€—^((βˆ’1)/( 2))βˆ’π‘₯ (γ€–1βˆ’π‘₯γ€—^2 )^((βˆ’1)/( 2)) = 0 = RHS 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.