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

Transcript

Ex 9.4, 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 ∫1▒〖𝑑𝑦 γ€—= ∫1β–’γ€–sin^(βˆ’1)⁑〖π‘₯.1 𝑑π‘₯γ€— γ€— y = sinβˆ’1 x ∫1β–’γ€–1 𝑑π‘₯ βˆ’βˆ«1β–’[1/√(1 βˆ’ π‘₯^2 ) ∫1β–’γ€–1.𝑑π‘₯ γ€—] γ€— dx 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 ("∴" (𝑑〖(sinγ€—^(βˆ’1) π‘₯))/𝑑π‘₯=1/√(1 βˆ’ π‘₯^2 )) y = x 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯ βˆ’ ∫1β–’π‘₯/√(1 βˆ’ π‘₯^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▒〖𝑑^((βˆ’1)/2) 𝑑𝑑〗 y = x sinβˆ’1 x + 1/2 𝑑^((βˆ’1)/2 + 1)/((βˆ’1)/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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.