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Ex 9.4, 9 - Find general solution: dy/dx = sin-1 x - Chapter 9

Ex 9.4, 9 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 9 - Chapter 9 Class 12 Differential Equations - Part 3

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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 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.