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

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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.