Ex 9.6, 13 - Find particular solution: dy/dx + 2y tan x = sin x - Ex 9.6

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Ex 9.6, 13 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : 𝑑𝑦/𝑑π‘₯+2𝑦 tan⁑〖π‘₯=sin⁑〖π‘₯;𝑦=0γ€— γ€— when π‘₯= πœ‹/3 𝑑𝑦/𝑑π‘₯+2𝑦 tan⁑〖π‘₯=sin⁑π‘₯ γ€— Differential equation is of the form 𝑑𝑦/𝑑π‘₯ + Py = Q 𝑑𝑦/𝑑π‘₯ + 2y tan x = sin x Where P = 2 tan x & Q = sin x Finding Integrating factor IF = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€— IF = 𝑒^∫1β–’γ€–2 tan⁑π‘₯ 𝑑π‘₯γ€— IF = e2 log sec x IF = 𝑒^log⁑sec^2⁑π‘₯ IF = sec2 x Solution is y (IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝑐〗 y (sec2 x) = ∫1β–’γ€–sin⁑π‘₯ sec^2⁑π‘₯ 𝑑π‘₯+𝑐〗 y sec2 x = ∫1β–’γ€–sin⁑π‘₯ 1/cos^2⁑π‘₯ γ€— dx + C y sec2 x = ∫1β–’γ€–sin⁑π‘₯/π‘π‘œπ‘ β‘π‘₯ Γ—1/π‘π‘œπ‘ β‘π‘₯ γ€— dx + C y sec2 x = ∫1β–’tan⁑〖π‘₯ sec⁑〖π‘₯ γ€— γ€— dx + C y sec2 x = sec⁑"x + C " y = sec⁑〖π‘₯ γ€—/sec^2⁑π‘₯ + 𝑐/sec^2⁑π‘₯ y = cos x + C cos2 x Putting x = πœ‹/3 & y = 0 0 = cos πœ‹/3 + C cos2 πœ‹/3 0 = 1/2 + C (1/2)^2 (βˆ’1)/2 = C (1/4) (βˆ’4)/2 = C C = βˆ’2 Putting value of C in (1) y = cos x + C cos2 x y = cos x βˆ’ 2 cos2 x

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