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Ex 9.3, 10 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑒^π‘₯ tan⁑〖𝑦 𝑑π‘₯+(1βˆ’π‘’^π‘₯ ) sec^2⁑〖𝑦 𝑑𝑦=0γ€— γ€— 𝑒^π‘₯ tan⁑〖𝑦 𝑑π‘₯+(1βˆ’π‘’^π‘₯ ) sec^2⁑〖𝑦 𝑑𝑦=0γ€— γ€— 𝑒^π‘₯ tan⁑〖𝑦 𝑑π‘₯=βˆ’(1βˆ’π‘’^π‘₯ ) sec^2⁑〖𝑦 𝑑𝑦〗 γ€— 𝑒^π‘₯ tan⁑〖𝑦 𝑑π‘₯γ€—=(𝑒^π‘₯βˆ’1) sec^2⁑〖𝑦 𝑑𝑦〗 𝒆^𝒙/(𝒆^𝒙 βˆ’ 𝟏) dx = (π’”π’†π’„πŸπ’š π’…π’š)/π’•π’‚π’β‘π’š Integrating both sides. ∫1▒〖𝑒^π‘₯/(𝑒^π‘₯ βˆ’ 1) 𝑑π‘₯γ€— = ∫1β–’γ€–(𝑠𝑒𝑐2 𝑦)/tan⁑𝑦 𝑑𝑦〗 Put 𝒆^π’™βˆ’πŸ = u and put tan y = v Diff u w.r.t. x ex = 𝑑𝑒/𝑑π‘₯ dx = 𝒅𝒖/𝒆𝒙 Diff v w.r.t. y sec2 y = 𝑑𝑣/𝑑𝑦 dy = 𝒅𝒗/γ€–π¬πžπœγ€—^πŸβ‘π’š Therefore ∫1▒〖𝑒π‘₯/𝑒 𝑑𝑒/𝑒π‘₯γ€— = ∫1▒𝑠𝑒𝑐2𝑦/(𝑣 𝑠𝑒𝑐2𝑦) dv ∫1▒𝑑𝑒/𝑒 = ∫1▒𝑑𝑣/𝑣 log u + c1 = log v Putting back u = ex βˆ’ 1 and V = tan y log |"ex βˆ’ 1" | + c1 = log tan y Putting c1 = log c log |"ex βˆ’ 1" |+ log c = log (tan y) log |𝑐("ex βˆ’ 1" )|= log |tan⁑𝑦 | 𝑐("ex βˆ’ 1" ) = tan y tan y = c ("ex βˆ’ 1" ) is the general solution

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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.