Check sibling questions


Transcript

Ex 9.3, 9 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑑𝑦/𝑑π‘₯=sin^(βˆ’1)⁑π‘₯𝑑𝑦/𝑑π‘₯=sin^(βˆ’1)⁑π‘₯ 𝑑𝑦 = sin^(βˆ’1)⁑π‘₯ dx Integrating both sides ∫1β–’γ€–π’…π’š γ€—= ∫1▒〖〖𝐬𝐒𝐧〗^(βˆ’πŸ)⁑〖𝒙.𝟏 𝒅𝒙〗 γ€— Integrating by parts, using formula ∫1▒〖𝑓 (π‘₯)𝑔(π‘₯)𝑑π‘₯ γ€—= 𝑓(π‘₯) ∫1▒〖𝑔(π‘₯)𝑑π‘₯ βˆ’βˆ«1β–’γ€–[𝑓′(π‘₯)∫1▒𝑔(π‘₯)𝑑π‘₯] 𝑑π‘₯ γ€— γ€— Take f(x) = sinβˆ’1 x and g(x) = 1 y = x γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) 𝒙 βˆ’ ∫1▒𝒙/√(𝟏 βˆ’ 𝒙^𝟐 ) dx Let t = 1 βˆ’ x2 dt = βˆ’2xdx x dx = (βˆ’π‘‘π‘‘)/2 Hence, our equation becomes y = x sinβˆ’1 x βˆ’ ∫1β–’(βˆ’π‘‘π‘‘)/(2βˆšπ‘‘) y = x sinβˆ’1 x + ∫1▒𝑑𝑑/(2βˆšπ‘‘) y = x sinβˆ’1 x + 𝟏/𝟐 ∫1▒〖𝒕^((βˆ’πŸ)/𝟐) 𝒅𝒕〗 y = x sinβˆ’1 x + 𝟏/𝟐 𝒕^((βˆ’πŸ)/𝟐 + 𝟏)/((βˆ’πŸ)/𝟐 + 𝟏) + C y = x sinβˆ’1 x + 1/2 (𝑑^(1/2) )/((1/2) )+𝐢 y = x sinβˆ’1 x + βˆšπ‘‘ + C Putting back value of t y = x sinβˆ’1 x + √(πŸβˆ’π’™^𝟐 ) + C y = sinβˆ’1 x ∫1β–’γ€–πŸ 𝒅𝒙 βˆ’βˆ«1β–’[𝟏/√(𝟏 βˆ’ 𝒙^𝟐 ) ∫1β–’γ€–πŸ.𝒅𝒙 γ€—] γ€— dx

  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

About the Author

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