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Ex 9.5, 4 For each of the differential equation given in Exercises 1 to 12, find the general solution : 𝑑𝑦/𝑑π‘₯+(sec⁑π‘₯ )𝑦=π‘‘π‘Žπ‘›π‘₯(0≀π‘₯<πœ‹/2) Differential equation is of the form 𝑑𝑦/𝑑π‘₯ + Py = Q where P = sec x and Q = tan x Finding integrating factor, IF = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€— IF = e^∫1β–’sec⁑〖π‘₯ 𝑑π‘₯γ€— IF = γ€–e^π‘™π‘œπ‘”γ€—^|sec⁑〖π‘₯ + tan⁑π‘₯ γ€— | I.F = sec x + tan x Solution is y (IF) = ∫1β–’(𝑄×𝐼.𝐹) 𝑑π‘₯+𝑐 y (sec x + tan x) = ∫1β–’γ€–π­πšπ§β‘π’™ (𝒔𝒆𝒄⁑𝒙+π­πšπ§β‘γ€–π’™)γ€— γ€—+𝒄 y (sec x + tan x) = ∫1β–’γ€–tan⁑π‘₯ sec⁑π‘₯ γ€— 𝑑π‘₯+∫1β–’γ€–tan^2⁑π‘₯ 𝑑π‘₯+𝐢〗 y (sec x + tan x) = sec x + ∫1β–’γ€–(sec^2⁑π‘₯βˆ’1)γ€— dx + c y (sec x + tan x) = sec x + π­πšπ§β‘π’™ βˆ’ x + c

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