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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.