Ex 9.3, 9 - Find general solution: dy/dx = sin-1 x - Chapter 9 - Ex 9.3

part 2 - Ex 9.3, 9 - Ex 9.3 - Serial order wise - Chapter 9 Class 12 Differential Equations
part 3 - Ex 9.3, 9 - Ex 9.3 - Serial order wise - Chapter 9 Class 12 Differential Equations

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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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